The Congruences of a Finite Lattice: A Proof-by-Picture Approach

George Gr¨atzer The Congruences of a Finite Lattice A Proof-by-Picture Approach Birkh¨auser Boston • Basel • Berlin ...

Author: George Grätzer

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George Gr¨atzer

The Congruences of a Finite Lattice A Proof-by-Picture Approach

Birkh¨auser Boston • Basel • Berlin

George Gr¨atzer University of Manitoba Department of Mathematics Winnipeg, MB, R3T 2N2 Canada

Cover design by Mary Burgess. AMS Subject Classiﬁcation: 06B10, 06D05, 06C05, 06C10

Library of Congress Control Number: 2005934394 ISBN 0-8176-3224-7 ISBN-13 978-0-8176-3224-3

eISBN 0-8176-4462-8

Printed on acid-free paper.

c 2006 Birkh¨auser Boston

All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Birkh¨auser Boston, c/o Springer Science+Business Media Inc., 233 Spring Street, New York, NY 10013, USA) and the author, except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks and similar terms, even if they are not identiﬁed as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed in the United States of America. 987654321 www.birkhauser.com

(EB)

To L´ aszl´ o Fuchs, my thesis advisor, my teacher, who taught me to set the bar high; and to the coauthors of my papers, Tomi (E. T. Schmidt), Harry (H. Lakser), Ervin (E. Fried), David (D. Kelly), Bob (R. W. Quackenbush), Fred (Wehrung), and more than ﬁfty others, who helped me raise the bar.

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii Glossary of Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xix Picture Gallery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxiii Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxv

I

A Brief Introduction to Lattices

1

1

Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Ordering. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.2 Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.3 Order constructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.4 Partitions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Lattices and semilattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Lattices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 Semilattices and closure systems . . . . . . . . . . . . . . . . . . . . . . . 1.3 Some algebraic concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.2 Sublattices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.3 Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 3 3 5 5 6 8 8 10 12 12 13 14

2

Special Concepts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.1 Elements and lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.2 Direct and subdirect products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.3 Polynomials and identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.4 Gluing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

viii

Contents

2.5

Modular and distributive lattices. . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 The characterization theorems . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Finite distributive lattices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.3 Finite modular lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30 30 31 32

3

Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.1 Congruence spreading. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.2 Prime intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.3 Congruence-preserving extensions and variants . . . . . . . . . . . . . . 39

II

Basic Techniques

4

Chopped Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 4.1 Basic deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 4.2 Compatible vectors of elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 4.3 Compatible vectors of congruences . . . . . . . . . . . . . . . . . . . . . . . . . . 50 4.4 From the chopped lattice to the ideal lattice . . . . . . . . . . . . . . . . 52 4.5 Sectional complementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5

Boolean Triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 5.1 The general construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 5.2 The congruence-preserving extension property . . . . . . . . . . . . . . 60 5.3 The distributive case. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 5.4 Two interesting intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

6

Cubic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 6.1 The construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 6.2 The basic property. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

III

Representation Theorems

45

77

7

The 7.1 7.2 7.3 7.4 7.5

Dilworth Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 The representation theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 Proof-by-Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 Computing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 Sectionally complemented lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

8

Minimal Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 8.1 The results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 8.2 Proof-by-Picture for minimal construction . . . . . . . . . . . . . . . . . . . 94 8.3 The formal construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 8.4 Proof-by-Picture for minimality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

Contents

8.5 8.6

ix

Computing minimality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

9

Semimodular Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 The representation theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Proof-by-Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Construction and proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

105 105 106 107 114

10

Modular Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 The representation theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Proof-by-Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Construction and proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

115 115 116 120 125

11

Uniform Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 The representation theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Proof-by-Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 The lattice N (A, B) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Formal proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

129 129 129 132 137 139

IV

Extensions

143

12

Sectionally Complemented Lattices . . . . . . . . . . . . . . . . . . . . . 12.1 The extension theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Proof-by-Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Simple extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 Formal proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

145 145 146 148 150 152

13

Semimodular Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 The extension theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Proof-by-Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 The conduit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4 The construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5 Formal proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

153 153 153 156 157 159 159

14

Isoform Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1 The result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Proof-by-Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3 Formal construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 The congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

161 161 161 165 171

x

Contents

14.5 The isoform property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 14.6 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 15

Independence Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Proof-by-Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2.1 Frucht lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2.2 An automorphism-preserving simple extension . . . . . . . . . 15.2.3 A congruence-preserving rigid extension . . . . . . . . . . . . . . . . 15.2.4 Merging the two extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2.5 The representation theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3 Formal proofs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3.1 An automorphism-preserving simple extension . . . . . . . . . 15.3.2 A congruence-preserving rigid extension . . . . . . . . . . . . . . . . 15.3.3 Proof of the independence theorems . . . . . . . . . . . . . . . . . . . . 15.4 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

177 177 178 178 179 180 181 182 183 183 185 185 187

16

Magic Wands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Constructing congruence lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1.1 Bijective maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1.2 Surjective maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Proof-by-Picture for bijective maps. . . . . . . . . . . . . . . . . . . . . . . . . . 16.3 Veriﬁcation for bijective maps. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 2/3-boolean triples. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.5 Proof-by-Picture for surjective maps . . . . . . . . . . . . . . . . . . . . . . . . 16.6 Veriﬁcation for surjective maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.7 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

189 189 189 190 191 194 198 204 206 207

V

Two Lattices

213

17

Sublattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1 The results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2 Proof-by-Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 Multi-coloring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.4 Formal proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.5 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

215 215 217 219 220 221

18

Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.1 The results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Proof-by-Picture for the main result . . . . . . . . . . . . . . . . . . . . . . . . . 18.3 A very formal proof: Main result. . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3.1 Categoric preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3.2 From DI to OR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

227 227 228 230 230 232

Contents

19

xi

18.3.3 From OR to HE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3.4 From CH to DI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3.5 From HE to CH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3.6 From CH to LA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3.7 The ﬁnal step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.4 Proof for sectionally complemented lattices . . . . . . . . . . . . . . . . . 18.5 Proof-by-Picture for planar lattices . . . . . . . . . . . . . . . . . . . . . . . . . . 18.6 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

232 233 234 237 237 238 241 242

Tensor Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.1 The problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.2 Three unary functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.3 Deﬁning tensor extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.4 Computing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.4.1 Some special elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.4.2 An embedding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.4.3 Distributive lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.5 Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.5.1 Congruence spreading. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.5.2 Some structural observations . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.5.3 Lifting congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.5.4 The main lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.6 The congruence isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.7 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

245 245 246 248 250 250 252 253 254 254 257 259 261 262 263

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275

Preface

The topic The congruences of a ﬁnite lattice L form a lattice, called the congruence lattice of L and denoted by Con L. The lattice Con L is a ﬁnite distributive lattice—according to a 1942 result of Funayama and Nakayama [21]. The converse is a result of R. P. Dilworth from 1944 (see [9]): Dilworth Theorem. Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnite lattice L. This result was ﬁrst published in 1962 in Gr¨ atzer and Schmidt [57]. In the 60 years since the discovery of this result, a large number of papers have been published, strengthening and generalizing the Dilworth Theorem. These papers form two distinct ﬁelds: (i) Representation theorems of ﬁnite distributive lattices as congruence lattices of lattices with special properties. (ii) The Congruence Lattice Problem (CLP): Can congruence lattices of lattices be characterized as distributive algebraic lattices? A nontrivial ﬁnite distributive lattice D is determined by the order J(D) of join-irreducible elements. So a representation of D as the congruence lattice of a ﬁnite lattice L is really a representation of a ﬁnite order P (= J(D)), as the order of join-irreducible congruences of a ﬁnite lattice L. A join-irreducible congruence of a nontrivial ﬁnite lattice L is exactly the same as a congruence of the form con(a, b), where a ≺ b in L; that is, the smallest congruence collapsing a prime interval. Therefore, it is enough to concentrate on such congruences, and make sure that they are ordered as required by P . The inﬁnite case is much diﬀerent. There are really only two general positive results: 1. The ideal lattice of a distributive lattice with zero is the congruence lattice of a lattice—see Schmidt [109] (also Pudl´ ak [100]).

xiv

Preface

2. Any distributive algebraic lattice with at most ℵ1 compact elements is the congruence lattice of a lattice—Huhn [92] and [93] (see also Dobbertin [13]). The big breakthrough for negative results came in 1999 in Wehrung [122], based on the results of Wehrung [121]. There is a recent survey paper of this ﬁeld by T˚ uma and Wehrung [119]. This book deals with the ﬁnite case.

The two types of representation theorems The basic representation theorems are all of the same general type. We represent a ﬁnite distributive lattice D as the congruence lattice of a “nice” ﬁnite lattice L. For instance, in the 1962 paper (Gr¨ atzer and Schmidt [57]), we had already proved that the ﬁnite lattice L for the Dilworth Theorem can be constructed as a sectionally complemented lattice. To understand the second—the more sophisticated—type of representation theorem, we need the concept of a congruence-preserving extension. Let L be a lattice, and let K be a sublattice of L. In general, there is not much connection between the congruence lattice of L and the congruence lattice of K. If they happen to be naturally isomorphic, we call L a congruence-preserving extension of K. (More formally, we require that the restriction map be an isomorphism, see Section 3.3.) For sectionally complemented lattices, the congruence-preserving extension theorem was published in a 1999 paper, Gr¨ atzer and Schmidt [69]: Every ﬁnite lattice K has a ﬁnite, sectionally complemented, congruence-preserving extension L. It is diﬃcult, reading this for the ﬁrst time, to appreciate how much stronger this theorem is than the straight representation theorem. While the 1962 theorem provides, for a ﬁnite distributive lattice D, a ﬁnite sectionally complemented lattice L whose congruence lattice is isomorphic to D, the 1999 theorem starts with an arbitrary ﬁnite lattice K, and builds a sectionally complemented lattice L on it with the same congruence structure.

Proof-by-Picture Trying to prove the Dilworth Theorem (unpublished at the time) in 1960 with Schmidt, we came up with the construction—more or less—as presented in Section 7.2. In 1960, we did not discover the 1968 result of Gr¨ atzer and Lakser [34] establishing that the construction of the chopped lattice solves the problem. So we translated the chopped lattice construction to a closure space, as in Section 7.4, proved that the closed sets form a sectionally complemented lattice L, and based on that, we veriﬁed that the congruence lattice of L represents the given ﬁnite distributive lattice. When we submitted the paper [57] for publication, it had a three-page section explaining the chopped lattice construction and its translation to a

Preface

xv

closure space. The referee was strict: “You cannot have a three-page explanation for a two-page proof.” I believe that in the 40 plus years since the publication of that article, few readers have developed an understanding of the idea behind the published proof. The referee’s dictum is quite in keeping with mathematical tradition and practice. When mathematicians discuss new results, they explain the constructions and the ideas with examples; when these same results are published, the motivation and the examples are largely gone. We publish deﬁnitions, constructions, and formal proofs (and conjectures, Paul Erd˝ os would have added). Tradition has it, when Gauss proved one of his famous results, he was not ready to publicize it because the proof gave away too much as to how the theorem was discovered. “I have had my results for a long time: but I do not yet know how I am to arrive at them”, Gauss is quoted in Arber [2]. In this book I try to break with this tradition. In most chapters, after stating the main result, I include a section: Proof-by-Picture. This is a misnomer. A Proof-by-Picture is not a proof. The Pythagorean Theorem has many well known Proofs-by-Picture—sometimes called “Visual Proofs”; these are really proofs. My Proof-by-Picture is an attempt to convey the idea of the proof. I trust that if the idea is properly understood, the reader should be able to provide the formal proof, or should at least have less trouble reading it. Think of a Proof-by-Picture as a lecture to an informed audience, concluding with “the formal details now you can provide.” I converted many of these sections into lectures; the transparencies for these can be found on my Web site: http://www.maths.umanitoba.ca/homepages/gratzer.html see the directory /MathBooks/lectures.html I will use the same Web site to post corrections to this book, problems solved, and so on.

Outline In the last paragraph, I call an audience “informed” if they are familiar with the basic concepts and techniques of lattice theory. Part I provides this. I am very selective as to what to include. Also, there are no proofs in this part— with a few exceptions—they are easy enough for the reader to work them out on his own. For proofs, lots of exercises, and a more detailed exposition, I refer the reader to my book [26]. (See also Davey and Priestley [11].) Most of the research in this book deals with representation theorems; lattices with certain properties are constructed with prescribed congruence structures. The constructions are ad hoc. Nevertheless, there are three basic techniques to prove representation theorems:

xvi

Preface

• chopped lattices, used in almost every chapter; • boolean triples, used in Chapters 10, 12, and 16, and generalized in Chapter 19; also used in some papers that did not make it in this book, for instance, Gr¨ atzer and Schmidt [71]; • cubic extensions, used in most chapters of Part IV. These are presented in Part II with proofs. Actually, there are two more basic techniques. Multi-coloring is used in three relevant papers: Gr¨ atzer, Lakser, and Schmidt [46], [49] and Gr¨ atzer and Schmidt [68]; however, it appears in the book only in Chapter 17, so we introduce it there. Pruning is utilized in Chapters 11 and 14—it would seem to qualify for Part II; however, there is no theory of pruning, just concrete uses, so there is no general theory to discuss in Part II. Part III contains the representation theorems, requiring only chopped lattices from Part II. I cover the following topics: • The Dilworth Theorem and the representation theorem for sectionally complemented lattices in Chapter 7 (Gr¨ atzer and Schmidt [57], Crawley and Dilworth [10]; see also [9]). • Minimal representations in Chapter 8; that is, for a given |J(D)|, we minimize the size of L representing D (Gr¨ atzer, Lakser, and Schmidt [45], Gr¨ atzer, Rival, and Zaguia [54]). • The semimodular representation theorem in Chapter 9 (Gr¨ atzer, Lakser, and Schmidt [48]). • The representation theorem for modular lattices in Chapter 10 ( Schmidt [106] and Gr¨ atzer and Schmidt [74]); we are forced to represent with a countable lattice L, since the congruence lattice of a ﬁnite modular lattice is always boolean. • The representation theorem for uniform lattices (that is, lattices in which any two congruence classes of a congruence are of the same size) in Chapter 11 (Gr¨ atzer, Schmidt, and Thomsen [79]). Part IV is mostly about congruence-preserving extension. I present the congruence-preserving extension theorem for • sectionally complemented lattices in Chapter 12 (Gr¨ atzer and Schmidt [69]); • semimodular lattices in Chapter 13 (Gr¨ atzer and Schmidt [72]); • isoform lattices (that is, lattices in which any two congruence classes of a congruence are isomorphic) in Chapter 14 (Gr¨ atzer, Quackenbush, and Schmidt [53]).

Preface

xvii

These three constructions are based on cubic extensions, introduced in Part II. In Chapter 15, I present the congruence-preserving extension version of the Baranski˘ı-Urquhart Theorem (Baranski˘ı [3], [4] and Urquhart [120]) on the independence of the congruence lattice and the automorphism group of a ﬁnite lattice (Gr¨ atzer and Schmidt [66]). Finally, in Chapter 16, I discuss two congruence “destroying” extensions, which we call “magic wands.” It is hoped that these can be used to construct new classes of algebraic distributive lattices as congruence lattices of lattices (Gr¨ atzer and Schmidt [75], Gr¨ atzer, Greenberg, and Schmidt [32]). What happens if we consider the congruence lattices of two lattices? I take up three variants of this question in Part V. Let L be a ﬁnite lattice, and let K be a sublattice of L. As we discuss it in Section 3.3, there is a map ext from Con K into Con L: For a congruence relation Θ of K, let the image ext Θ be the congruence relation conL (Θ) of L generated by Θ. The map ext is a {0}-separating join-homomorphism. Chapter 17 proves the converse, a 1974 result of Huhn [91] and a stronger form due to Gr¨ atzer, Lakser, and Schmidt [46]. In Chapter 18, we deal with ideals. Let I be an ideal of a lattice L. Then the restriction map re : Con L → Con I (which assigns to a congruence Θ of L, the restriction ΘK of Θ to K) is a {0, 1}-homomorphism. We prove the corresponding representation theorem for ﬁnite lattices—Gr¨ atzer and Lakser [35]. We also prove two variants. The ﬁrst is by Gr¨ atzer and Lakser [42] that this result also holds for sectionally complemented lattices. The second is by Gr¨ atzer and Lakser [40] that this result also holds for planar lattices. The ﬁnal chapter is a ﬁrst contribution to the following class of problems. Let be a construction for ﬁnite distributive lattices (that is, if D and E are ﬁnite distributive lattices, then so is D E). Find a construction of ﬁnite lattices (that is, if K and L are ﬁnite lattices, then so is K L) satisfying Con(K L) ∼ = Con K Con L. If is the direct product, the answer is obvious since Con(K × L) ∼ = Con K × Con L. In Chapter 19, we take up the construction D E = D[E], deﬁned as the distributive lattice of all isotone maps from J(E) to D. In Gr¨ atzer and Greenberg [29], we introduced a construction: the tensor extension A[B], for nontrivial ﬁnite lattices A and B. In Chapter 19, we prove that Con(A[B]) ∼ = (Con A)[Con B]. The background of this result is described in Chapter 19. Each chapter in Parts III–V concludes with an extensive discussion section, giving the background for the topic, further results, and open problems. This book lists almost 80 open problems, hoping to convince the reader that we have hardly started. There are also more than 120 references and a detailed index of about 1500 entries.

xviii

Preface

This book is, as much as possible, visually oriented. I cannot stress too much the use of diagrams as a major research tool in lattice theory. I did not include in the book the list of ﬁgures because there is not much use to it; it lists over 110 ﬁgures. Notation and terminology Lattice-theoretic terminology and notation evolved from the three editions of G. Birkhoﬀ’s Lattice Theory; [8], by way of my books, [22]–[26], and McKenzie, McNulty, and Taylor [97], changing quite a bit in the process. Birkhoﬀ’s notation for the congruence lattice and ideal lattice of a lattice changed from Θ(L) and I(L) to Con L and Id L, respectively. The advent of LATEX promoted the use of operators for lattice constructions. I try to be consistent: I use an operator when a new structure is constructed; so I use Con L, Id L, Aut L, and so on, without parentheses, unless required for readability, for instance, J(D) and Con(Id L). I use functional notation when sets are constructed, as in Atom(L) and J(a). “Generated by” uses the same letters as the corresponding lattice construction, but starting with a lower case letter: con(H) is the congruence generated by H and id(H) is the ideal generated by H. I reversed the arrows for projectivities as compared with my book [26]. I think the new notation is easier to remember: congruences spread as the arrows point. This is also more consistent with the use in universal algebra. New concepts introduced in more recent research papers exhibit the usual richness in notation and terminology. I use this opportunity, with the wisdom of hindsight, to make their use consistent. The reader will often ﬁnd diﬀerent notation and terminology when reading the original papers. The detailed Glossary of Notation and Index may help. In combinatorial results, I will use Landau’s O notation: for the functions f and g, we write f = O(g) to mean that |f | ≤ C|g|, for a suitable constant C.

Winnipeg, Manitoba George Gr¨ atzer Summer, 2005 Homepage: http://www.maths.umanitoba.ca/homepages/gratzer.html/

Glossary of Notation

Symbol

Explanation

Page

Atom(U ) Aut L Bn Cn con(a, b) con(c) con(H) con(p) Con L ConJ L ConM L Cube K D Diag Down P ext : Con K → Con L fil(a) fil(H) FD (3) FK (H) FM (3) Frucht C hom{∨,0} (X, Y )

set of atoms of the ideal U 84 automorphism group of L 12 boolean lattice with n atoms 4 n-element chain 4 smallest congruence under which a ≡ b 15 principal congruence for a color c 39 smallest congruence collapsing H 16 principal congruence for the prime interval p 37 congruence lattice of L 15, 48 order of join-irreducible congruences of L 37 order of meet-irreducible congruences of L 71 cubic extension of K 71 class (variety) of distributive lattices 24 diagonal embedding of K into Cube K 71 order of down-sets of the (hemi)order P 4, 9, 232 for K ≤ L, extension map: Θ → conL (Θ) 41 ﬁlter generated by the element a 14 ﬁlter generated by the set H 14 free distributive lattice on three generators 26 free lattice generated by H in a variety K 26 free modular lattice on three generators 28 Frucht lattice of a graph C 178 {∨, 0}-homomorphism of X into Y 253

xx

Glossary of Notation

Symbol

Explanation

id(a) id(H) Id L (Id) Isoform J(D) J(ϕ) J(a) ker(ϕ) L M Max mcr(n) mcr(n, V) M(D) M3

ideal generated by the element a 14 ideal generated by the set H 14 ideal lattice of L 14, 48 condition to deﬁne ideals 14, 48 class of isoform lattices 141 order of join-irreducible elements of D 19 J(ϕ) : J(E) → J(D), the “inverse” of ϕ : D → E 32 set of join-irreducible elements below a 19 congruence kernel of ϕ 16 class (variety) of all lattices 25 class (variety) of modular lattices 25 maximal elements of an order 49 minimal congruence representation function 87 mcr for a class V 87 order of meet-irreducible elements of D 32 ﬁve-element modular nondistributive lattice xxiii, 11, 30 order of boolean triples of L 58 interval of M3 [L] 63 interval of M3 [L] 65 order of boolean triples of the interval [a, b] 58 reﬂection of Θ3 to M3 [L] 60 reﬂection of Θ3 to M3 [L, a] 64 reﬂection of Θ3 to M3 [L, a, b] xxiii, 67 ﬁve-element nonmodular lattice xxiii, 11, 30 seven-element nonmodular lattice 94 six-element nonmodular lattice xxiii, 80 2/3-boolean triple construction 198 lattice construction 132 Landau O notation xviii partition lattice of A 7, 9 power set lattice of X 4 order of nonempty subsets of X 219 set of prime intervals of L 37 reﬂection (restriction) map: Θ → ΘK 39 class of sectionally complemented lattices 87 class of semimodular lattices 87 simple extension of K 71 join-substitution property 14, 48 meet-substitution property xxiii, 14, 48 sublattice generated by H 13 eight-element semimodular lattice 106 class (variety) of trivial lattices 25 class of uniform lattices 141

M3 [L] M3 [L, a] M3 [L, a, b] M3 [a, b] M3 [Θ] M3 [Θ, a] M3 [Θ, a, b] N5 N5,5 N6 = N (p, q) N6 [L] N (A, B) O(f ) Part A Pow X Pow+ X Prime(L) re : Con L → Con K SecComp SemiMod Simp K (SP∨ ) (SP∧ ) sub(H) S8 T Uniform

Page

Glossary of Notation

Symbol

Explanation

xxi Page

Relations and Congruences A2 set of ordered pairs of A , τ , π, . . . binary relations Θ, Ψ, . . . congruences ω zero of Part A ι unit of Part A a ≡ b (π) a and b in the same block of π ab a and b in relation a ≡ b (Θ) a and b in relation Θ a/π block containing a H/π blocks represented by H α◦β product of α and β r α◦β reﬂexive product of α and β ΘK restriction of Θ to the sublattice K L/Θ quotient lattice Φ/Θ quotient congruence πi projection map: L1 × · · · × Ln → Li Θ×Φ direct product of congruences

7 7 7 3 3 6, 14 7 21 30 14 16 16 21 21

Orders ≤, < ≥, > K≤L ≤Q a b a≺b ba 0 1 a∨b H a ∧ b H Pd [a, b] ↓H ↓a P ∼ =Q

3 3 13 4 3 5 5 4 4 9 3 9 4 4, 10 13 4 4 4, 12

ordering ordering, inverse notation K a sublattice of L ordering of P restricted to a subset Q a incomparable with b a is covered by b b covers a zero, least element of an order unit, largest element of an order join operation least upper bound of H meet operation greatest lower bound of H dual of the order (lattice) P interval down-set generated by H down-set generated by {a} order (lattice) P isomorphic to Q

3

xxii

Glossary of Notation

Symbol

Explanation

Page

Constructions P ×Q P +Q P Q A[B] A⊗B U V

direct product of P and Q sum of P and Q glued sum of P and Q tensor extension of A by B tensor product of A and B modular lattice construction

5, 20 6 16 248 245 120

Perpectivities [a, b] ∼ [c, d] u [a, b] ∼ [c, d]

[a, b] perspective to [c, d] [a, b] up-perspective to [c, d]

32 33

[a, b] ∼ [c, d] [a, b] ≈ [c, d] [a, b] [c, d] [a, b] [c, d] [a, b] → [c, d] [a, b] ⇒ [c, d] [a, b] ⇔ [c, d]

[a, b] down-perspective to [c, d] [a, b] projective to [c, d] [a, b] up congruence-perspective onto [c, d] [a, b] down congruence-perspective onto [c, d] [a, b] congruence-perspective onto [c, d] [a, b] congruence-projective onto [c, d] [a, b] ⇒ [c, d] and [c, d] ⇒ [a, b]

33 33 35 35 35 36 36

Prime intervals p, q, . . . con(p) p⇒q p⇔q Prime(L)

principal congruence generated by p p is congruence-projective onto q p ⇒ q and q ⇒ p set of prime intervals of L

37 36 36 37

Miscellaneous x ∅

closure of x empty set

10 4

d

Picture Gallery

C3

C2

N5

S8

B2

N5,5

N6

M3

Acknowledgments

In 2002, I was invited to Nashville to a meeting. I wrote to E. T. Schmidt about the invitation. He could not come, but we agreed that I would give a survey lecture on ﬁnite congruence lattices. This was published in [78]. Later in the year, I was invited to the Summer School on General Algebra and Ordered Sets in Tale, Slovakia, to give a series of lectures on the same topic. Unfortunately, Schmidt did not attend. This the book is an outgrowth of my Tale lectures. Special thanks to the organizers of that excellent meeting: Miroslav Haviar, Tibor Katriˇ na´k, and Miroslav Ploˇsˇcica. Every chapter in Parts II–V is based on one or more joint papers I wrote with my coauthors. Without their collaboration, this book could not have existed. My seminar continues to correct errors and improve my presentation. In all matters notational and linguistic, David Kelly is my chief advisor. I got useful feedback from my fourth year honors algebra class, and especially from Mercedes Scott. I spent two months at the Institute of Advance Study at La Trobe University, Melbourne, Australia, where I had an opportunity to present all the Proof-by-Picture sections of this book at the seminar of Brian Davey. I would like to thank the IAS (Gilah Leder, Director, Julia Anderson, Executive Ofﬁcer and Michael Osborne, Vice Chancellor) for their wonderful hospitality and for the perfect work environment. Brian and his seminar recommended many improvements and corrected many mistakes. Special thanks to those who read (part of) my book and sent me comments. Kira Adaricheva suggested many improvements to the ﬁrst seven chapters. Fred Wehrung ﬁlled seven pages with objections minor and major; this is a much better book for his eﬀort.

Part I

A Brief Introduction to Lattices

1 Basic Concepts

In this chapter we introduce the most basic order theoretic concepts: orders, lattices, diagrams, and the most basic algebraic concepts: sublattices, congruences, products.

1.1.

Ordering

1.1.1

Orders

A binary relation on a nonempty set A is a subset of A2 , that is, a set of ordered pairs a, b, with a, b ∈ A. For a, b ∈ , we will write a b or a ≡ b (). A binary relation ≤ on a set P is called an ordering if it is reﬂexive (a ≤ a, for all a ∈ P ), antisymmetric (a ≤ b and b ≤ a imply that a = b, for all a, b ∈ P ), and transitive (a ≤ b and b ≤ c imply that a ≤ c, for all a, b, c ∈ P ). An order P, ≤ consists of a nonempty set P and an ordering ≤. a < b means that a ≤ b and a = b. We also use the “inverse” relations: a ≥ b deﬁned as b ≤ a and a > b for b < a. If more than one ordering is being considered, we write ≤P for the ordering of P, ≤; on the other hand if the ordering is understood, we will say that P (rather than P, ≤) is an order. An order P is trivial if P has only one element. The elements a and b of the order P are comparable if a ≤ b or b ≤ a. Otherwise, a and b are incomparable, in notation, a b. Let H ⊆ P and a ∈ P . Then a is an upper bound of H iff h ≤ a, for all h ∈ H. An upper bound a of H is the least upper bound of H iff, a ≤ b,for any upper bound b of H; in this case, we will write a = H. If a = H

4

1. Basic Concepts

exists, then it is unique. By deﬁnition, ∅ exist (∅ is the empty set) iff P has a smallest element, zero, denoted by 0. The concepts of lower bound and greatest lower bound are similarly deﬁned; the latter is denoted by H. Note that ∅ exists iff P has a largest element, unit, denoted by 1. A bounded order is one that has both 0 and 1. The adverb “similarly” (in “similarly deﬁned”) in the previous paragraph can be given concrete meaning. Let P, ≤ be an order. Then P, ≥ is also an order, called the dual of P, ≤. The dual of the order P will be denoted by P d . Now if Φ is a “statement” about orders, and if in Φ we replace all occurrences of ≤ by ≥, then we get the dual of Φ. Duality Principle for Orders. If a statement Φ is true for all orders, then its dual is also true for all orders. A chain (linear order ) is an order with no incomparable elements. An antichain is one in which a b, for all a = b. Let P, ≤ be an order and let Q be a nonempty subset of P . Then there is a natural order ≤Q on Q induced by ≤: for a, b ∈ Q, let a ≤Q b iff a ≤ b; we call Q, ≤Q (or simply, Q, ≤, or even simpler, Q) a suborder of P, ≤. A chain C in an order P is a nonempty subset, which, as a suborder, is a chain. An antichain C in an order P is a nonempty subset which, as a suborder, is an antichain. The length of a ﬁnite chain C, length C, is |C| − 1. An order P is said to be of length n (in formula, length P = n), where n is a natural number iff there is a chain in P of length n and all chains in P are of length ≤ n. The orders P and Q are isomorphic (in formula, P ∼ = Q) and the map ϕ : P → Q is an isomorphism iff ϕ is one-to-one and onto and a ≤ b in P

iff aϕ ≤ bϕ in Q.

Let Cn denote the set {0, . . . , n − 1} ordered by 0 < 1 < 2 < · · · < n − 1. Then Cn is an n-element chain. Observe that length Cn = n − 1. If C = {x0 , . . . , xn−1 } is an n-element chain and x0 < x1 < · · · < xn−1 , then ϕ : i → xi is an isomorphism between Cn and C. Therefore, the n-element chain is unique up to isomorphism. Let Bn denote the set of all subsets of the set {0, . . . , n − 1} ordered by containment. Observe that the order Bn has 2n elements and length Bn = n. In general, for a set X, we denote by Pow X the power set of X, that is, the set of all subsets of X ordered by set inclusion. For an order P , call A ⊆ P a down-set iff x ∈ A and y ≤ x imply that y ∈ A. For H ⊆ P , there is a smallest down-set containing H, namely, { x | x ≤ h, for some h ∈ H }; we use the notation ↓ H for this set. If H = {a}, we write ↓ a for ↓ {a}. Let Down P denote the set of all down-sets ordered by set inclusion. If P is an antichain, then Down P ∼ = Bn , where n = |P |. The map ϕ : P1 → P2 is an isotone map (resp., antitone map) of the order P1 into the order P2 iff a ≤ b in P1 implies that aϕ ≤ bϕ (resp., aϕ ≥ bϕ)

1.1. Ordering

5

in P2 . Then P1 ϕ is a suborder of P2 . Even if ϕ is one-to-one, the orders P1 and P2 ϕ need not be isomorphic. 1.1.2

Diagrams

In the order P , the element a is covered by b or b covers a (in formula, a ≺ b or b a) iff a < b and a < x < b, for no x ∈ P . The binary relation ≺ is called the covering relation. The covering determines the ordering: Let P be a ﬁnite order. Then a ≤ b iff a = b or if there exists a ﬁnite sequence of elements x1 , x2 , . . . , xn such that a = x1 ≺ x2 ≺ · · · ≺ xn = b. A diagram of an order P represents the elements with small circles ; the circles representing two elements x, y are connected by a line segment iff one covers the other; if x is covered by y, then the circle representing x is placed lower than the circle representing y. The diagram of a ﬁnite order determines the order up to isomorphism. In a diagram the intersection of two line segments does not indicate an element. A diagram is planar if no two line segments intersect. An order P is planar if it has a diagram that is planar. Figure 1.1 shows three diagrams of the same order P . Since the third diagram is planar, P is a planar order. 1.1.3

Order constructions

Given the orders P and Q, we can form the direct product P × Q, consisting of all ordered pairs x1 , x2 , with x1 ∈ P and x2 ∈ Q, ordered componentwise, that is, x1 , x2 ≤ y1 , y2 iff x1 ≤ y1 and x2 ≤ y2 . If P = Q, then we write P 2 for P × Q. Similarly, we use the notation P n for P n−1 × P , for n > 2. Figure 1.2 shows a diagram of C2 × P , where P is the order with diagrams in Figure 1.1.

Figure 1.1: Three diagrams of an order.

6

1. Basic Concepts

Figure 1.2: A diagram of C2 × P . Another often used construction is the (ordinal) sum P + Q of P and Q, deﬁned on the (disjoint) union P ∪ Q and ordered as follows: ⎧ ⎪ ⎨x ≤P y, for x, y ∈ P ; x ≤ y iff x ≤Q y, for x, y ∈ Q; ⎪ ⎩ x ∈ P, y ∈ Q. Figure 1.3 shows diagrams of C2 + P and P + C2 , where P is the order with diagrams in Figure 1.1. In both diagrams, the elements of C2 are black-ﬁlled. Figure 1.3 also shows diagram of P + C2 . A variant of this is the glued sum, P Q, applied to an order P with largest element 1P and an order Q with smallest element 0Q ; then P Q is P + Q in which 1P and 0Q are identiﬁed (that is, 1P = 0Q in P Q). 1.1.4

Partitions

We now give a nontrivial example of an order. A partition of a nonempty set A is a set π of nonempty pairwise disjoint subsets of A whose union is A. The members of π are called the blocks of π. The block containing a ∈ A will be denoted by a/π. A singleton as a block is called trivial. If the elements a and

Figure 1.3: Diagrams of C2 + P , P + C2 , and P C2 .

1.1. Ordering

7

b of A belong to the same block, we write a ≡ b (π) or a π b or a/π = b/π. In general, for H ⊆ A, H/π = { a/π | a ∈ H }. An equivalence relation ε on the set A is a reﬂexive, symmetric (aεb implies that bεa, for all a, b ∈ A), and transitive binary relation. Given a partition π, we can deﬁne an equivalence relation ε by x, y ∈ ε iff x/π = y/π. Conversely, if ε is an equivalence relation, then π = { a/ε | a ∈ A } is a partition of A. There is a one-to-one correspondence between partitions and equivalence relations; we will use the two terms interchangeably. Part A will denote the set of all partitions of A ordered by π1 ≤ π2

iff x ≡ y (π1 ) implies that x ≡ y (π2 ).

We draw a picture of a partition by drawing the boundary lines of the (nontrivial) blocks. Then π1 ≤ π2 iff the boundary lines of π2 are also boundary lines of π1 (but π1 may have some more boundary lines). Equivalently, the blocks of π2 are unions of blocks of π1 ; see Figure 1.4.

A

π1 ≤ π 2 and

π1 : π2 :

Figure 1.4: Drawing a partition. Part A has a zero and a unit, denoted by ω and ι, respectively, deﬁned by x≡y x≡y

(ω) (ι),

iff x = y; for all x, y ∈ A.

Figure 1.5 shows the diagrams of Part A, for |A| ≤ 4. The partitions are labeled by listing the nontrivial blocks. A preorder is a nonempty set Q with a binary relation ≤ that is reﬂexive and transitive. Let us deﬁne the binary relation a ≈ b on Q as a ≤ b and b ≤ a. Then ≈ is an equivalence relation. Deﬁne the set P as Q/≈, and on P deﬁne the binary relation ≤: a/≈ ≤ b/≈

iff a ≤ b in Q.

8

1. Basic Concepts ι = {1, 2, 3}

ι = {1, 2}

{1, 2}

{1, 3}

{2, 3}

Part {1, 2} Part {1}

ω

Part {1, 2, 3}

ω

{1, 2, 3, 4} = ι {1, 2, 3}

{2, 3, 4} {1, 3, 4}

{1, 2}, {3, 4}

{1, 2}

{1, 3}

{1, 4}, {2, 3}

{1, 3}, {2, 4}

{1, 4}

{2, 3}

ω

{2, 4}

{1, 2, 4}

{3, 4}

Part {1, 2, 3, 4}

Figure 1.5: Part A, for |A| ≤ 4. It is easy to see that the deﬁnition of ≤ on P is well-deﬁned and that P is an order. We will call P the order associated with the preorder Q. Starting with a binary relation ≺ on the set Q, we can deﬁne the reﬂexivetransitive closure ≤ of ≺ by the formula: for a, b ∈ Q, let a ≤ b iff a = b or if a = x0 ≺ x1 ≺ · · · ≺ xn = b, for elements x1 , . . . , xn−1 ∈ Q. Then ≤ is a preordering on Q. A cycle on Q is a sequence x1 , . . . , xn ∈ Q satisfying x1 ≺ x2 · · · ≺ xn ≺ x1 (n > 1). The preordering ≤ is an ordering iff there are no cycles.

1.2.

Lattices and semilattices

1.2.1

Lattices

An order L, ≤ is a lattice if {a, b} and {a, b} exist, for all a, b ∈ L. A lattice L is trivial if it has only one element; otherwise, it is nontrivial.

1.2. Lattices and semilattices

9

We will use the notations a∨b= a∧b=

{a, b}, {a, b},

and call ∨ the join and ∧ the meet. They are both binary operations that are idempotent (a ∨ a = a and a ∧ a = a), commutative (a ∨ b = b ∨ a and a ∧ b = b ∧ a), associative ((a ∨ b) ∨ c = a ∨ (b ∨ c) and (a ∧ b) ∧ c = a ∧ (b ∧ c)), and absorptive (a ∨ (a ∧ b) = a and a ∧ (a ∨ b) = a). These properties of the operations are also called the idempotent identities, commutative identities, associative identities, and absorption identities, respectively. (Identities, in general, are introduced in Section 2.3.) As always in algebra, associativity makes it possible to write a1 ∨ a2 ∨ · · · ∨ an without using parentheses (and the same for ∧). For instance, for A, B ∈ Pow X, we have A∨B = A∪B and A∧B = A∩B. So Pow X is a lattice. For Θ, Φ ∈ Part A, if we regard Θ and Φ as equivalence relations, then the meet formula is trivial: Θ ∧ Φ = Θ ∩ Φ, but the formula for joins is a bit more complicated: x ≡ y (Θ ∨ Φ) iff there is a sequence x = z0 , z1 , . . . , zn = y of elements of A such that zi ≡ zi+1 (Θ) or zi ≡ zi+1 (Φ), for each 0 ≤ i < n. So Part A is a lattice; it is called the partition lattice on A. For an order P , the order Down P is a lattice: A ∨ B = A ∪ B and A ∧ B = A ∩ B, for A, B ∈ Down P . An (n-ary) operation on a nonempty set A is a map from An to A. For n = 2, we call the operation binary. An algebra is a nonempty set A with operations deﬁned on A. To treat lattices as algebras, deﬁne an algebra L, ∨, ∧ a lattice iff L is a nonempty set, ∨ and ∧ are binary operations on L, both ∨ and ∧ are idempotent, commutative, and associative, and they satisfy the two absorption identities. A lattice as an algebra and a lattice as an order are “equivalent” concepts: Let the order L = L, ≤ be a lattice. Then the algebra La = L, ∨, ∧ is a lattice. Conversely, let the algebra L = L, ∨, ∧ be a lattice. Deﬁne a ≤ b iﬀ a ∨ b = b. Then Lp = L, ≤ is an order, and the order Lp is a lattice. For an order L that is a lattice, we have Lap = L; for an algebra L that is a lattice, we have Lpa = L. Note that for lattices as algebras, the Duality Principle takes on the following very simple form. Duality Principle for Lattices. Let Φ be a statement about lattices expressed in terms of ∨ and ∧. The dual of Φ is the statement we get from Φ by interchanging ∨ and ∧. If Φ is true for all lattices, then the dual of Φ is also true for all lattices.

10

1. Basic Concepts

If the operations are understood, we will say that L (rather than L, ∨, ∧) is a lattice. The dual of the lattice L will be denoted by Ld ; the order Ld is also a lattice. A ﬁnite lattice L is planar if it is planar as an order (see Section 1.1.2). We have quite a bit of ﬂexibility to construct a planar diagram for an order, but for a lattice, we are much more constrained because L has a zero, which must be the lowest element and a unit, which must be the highest element— contrast this with Figure 1.1. All lattices with ﬁve or fewer elements are planar; all but the ﬁve chains are shown in the ﬁrst two rows of Figure 1.6 (see next page). The third row of Figure 1.6 provides a good example of “good” and “bad” lattice diagrams; the two diagrams represent the same lattice, C32 . Planar diagrams are the best. Diagrams in which meets and joins are hard to ﬁgure out are not of much value. In the last row of Figure 1.6 there are two more diagrams. The one on the left is not planar; nevertheless, it is very easy to work with: meets and joins are easy to see (the notation M3 [C3 ] will be explained in Section 5.1). The one on the right is not a lattice: the two black-ﬁlled elements have no join. In this book we deal almost exclusively with ﬁnite lattices. Some concepts, however, are more natural to introduce in a more general context. An order L, ≤ is a complete lattice if X and X exist, for all X ⊆ L. All ﬁnite lattices are complete, of course. 1.2.2

Semilattices and closure systems

A join-semilattice S, ∨ is an algebra: a nonempty set S with an idempotent, commutative, and associative binary operation ∨. In a join-semilattice S, ∨, we can deﬁne an ordering: a ≤∨ b iff a ∨ b = b. In the order S, ≤∨ , we have {a, b} = a ∨ b. Similarly, a meet-semilattice S, ∧ is an algebra: a nonempty set S with an idempotent, commutative, and associative binary operation ∧. In a meetsemilattice S, ∧,we can deﬁne an ordering: a ≤∧ b iff a ∧ b = a. In the order S, ≤∧ , we have {a, b} = a ∧ b. If the operation is understood, we will say that S (rather than S, ∨) is a join-semilattice; similarly, for a meet-semilattice. If L, ∨, ∧ is a lattice, then L, ∨ is a join-semilattice and L, ∧ is a meet-semilattice; moreover, the orderings ≤∨ and ≤∧ agree. The converse also holds. Let L be a lattice and let C be a nonempty subset of L with the property that for every x ∈ L, there is a smallest element x of C with x ≤ x. We call C a closure system in L, and x the closure of x in C. Obviously, C, as a suborder of L, is a lattice: For x, y ∈ C, the meet in C is the same as the meet in L, and the join is x ∨C y = x ∨L y.

11

1.2. Lattices and semilattices

C1 + B2

B2

B2 + C1

i

i

a a

c

b

c

b

M3

N5 o

o

C32

M3 [C3 ] Figure 1.6: More diagrams.

12

1. Basic Concepts

Let L be a complete lattice and let C be -closed subset of L, that is, if X ⊆ C, then X ∈ C. (Since ∅ = 1, such a subset is nonempty and contains the 1 of L.) Then C is a closure system in L, and for every x ∈ L, x = ( y ∈ C | x ≤ y ).

1.3.

Some algebraic concepts

1.3.1

Homomorphisms

The lattices L1 = L1 , ∨, ∧ and L2 = L2 , ∨, ∧ are isomorphic as algebras (in symbols, L1 ∼ = L2 ), and the map ϕ : L1 → L2 is an isomorphism iff ϕ is one-to-one and onto and (1)

(a ∨ b)ϕ = aϕ ∨ bϕ,

(2)

(a ∧ b)ϕ = aϕ ∧ bϕ,

for a, b ∈ L1 . A map, in general, and a homomorphism, in particular, is called a surjection if it is onto, and a bijection if it is also one-to-one. An isomorphism of a lattice with itself is called an automorphism. The automorphisms of a lattice L form a group Aut L under composition. A lattice L is rigid if the identity map is the only automorphism of L, that is, if Aut L is the one-element group. It is easy to see that two lattices are isomorphic as orders iff they are isomorphic as algebras. Let us deﬁne a homomorphism of the join-semilattice S1 , ∨ into the joinsemilattice S2 , ∨ as a map ϕ : S1 → S2 satisfying (1); similarly, for meetsemilattices, we require (2). A lattice homomorphism (or simply, homomorphism) ϕ of the lattice L1 into the lattice L2 is a map of L1 into L2 satisfying both (1) and (2). A homomorphism of a lattice into itself is called an endomorphism. A one-to-one homomorphism will also be called an embedding. Note that meet-homomorphisms, join-homomorphisms, and (lattice) homomorphisms are all isotone.

Figure 1.7: Morphisms. Figure 1.7 shows three maps of the four-element lattice B2 into the threeelement chain C3 . The ﬁrst map is isotone but it is neither a meet- nor a

1.3. Some algebraic concepts

13

join-homomorphism. The second map is a join-homomorphism but is not a meet-homomorphism, thus not a homomorphism. The third map is a (lattice) homomorphism. Various versions of homomorphisms and embeddings will be used. For instance, for lattices and join-semilattices, there are also {∨, 0}-homomorphism, and so on, with obvious meanings. An onto homomorphism ϕ is also called surjective, while a one-to-one homomorphism is called injective; it is the same as an embedding. For bounded lattices, we often use {0, 1}-homomorphisms and {0, 1}-embeddings. It should always be clear from the context what kind of homomorphism we are considering. If we say, “let ϕ be a homomorphism of K into L”, where K and L are lattices, then ϕ is a lattice homomorphism, unless otherwise stated. 1.3.2

Sublattices

A sublattice K, ∨, ∧ of the lattice L, ∨, ∧ is deﬁned on a nonempty subset K of L with the property that a, b ∈ K implies that a ∨ b, a ∧ b ∈ K (the operations ∨, ∧ are formed in L, ∨, ∧), and the ∨ and the ∧ of K, ∨, ∧ are restrictions to K of the ∨ and the ∧ of L, ∨, ∧, respectively. Instead of “K, ∨, ∧ is a sublattice of L, ∨, ∧”, we will simply say that “K is a sublattice of L”—in symbols, K ≤ L. Of course, a sublattice of a lattice is again a lattice. If K is a sublattice of L, then we call L an extension of K—in symbols, L ≥ K. For a bounded lattice L, the sublattice K is a {0, 1}-sublattice if K ≤ L and the 0 and 1 of L are in K. Similarly, we can deﬁne a {0}-sublattice, and so on. For every H ⊆ L, H = ∅, there is a smallest sublattice sub(H) ⊆ L containing H called the sublattice of L generated by H. We say that H is a generating set of sub(H). The subset K of the lattice L is called convex iff a, b ∈ K, c ∈ L, and a ≤ c ≤ b imply that c ∈ K. We can add the adjective “convex” to sublattices, extensions, and embeddings. A sublattice K of the lattice L is convex if it a convex subset of L. Let L be an extension of K; then L is a convex extension if K is a convex sublattice. An embedding is convex if the image is a convex sublattice. For a, b ∈ L, a ≤ b, the interval [a, b] = { x | a ≤ x ≤ b } is an important example of a convex sublattice. An interval [a, b] is trivial if a = b. The smallest nontrivial intervals are called prime; that is, [a, b] is prime iff a ≺ b. Another important example of a convex sublattice is an ideal. A nonempty subset I of L is an ideal iff it is a down-set with the property: (Id) a, b ∈ I implies that a ∨ b ∈ I.

14

1. Basic Concepts

An ideal I of L is proper if I = L. Since the intersection of any number of ideals is an ideal, unless empty, we can deﬁne id(H), the ideal generated by a subset H of the lattice L, provided that H = ∅. If H = {a}, we write id(a) for id({a}), and call it a principal ideal. Obviously, id(a) = { x | x ≤ a } =↓ a. So instead of id(a), we could use ↓ a; many do, who work in categorical aspects of lattice theory—and use id for the identity map. The set Id L of all ideals of L is an order under set inclusion, and as an order it is a lattice. In fact, for I, J ∈ Id L, the lattice operations in Id L are I ∨ J = id(I ∪ J) and I ∧ J = I ∩ J. So we obtain the formula for the ideal join: x ∈ I ∨ J iff x ≤ i ∨ j, for some i ∈ I, j ∈ J. We call Id L the ideal lattice of L. Now observe the formulas: id(a) ∨ id(b) = id(a ∨ b), id(a) ∧ id(b) = id(a ∧ b). Since a = b implies that id(a) = id(b), these yield: The map a → id(a) embeds L into Id L. Since the deﬁnition of an ideal uses only ∨ and ≤, it applies to any joinsemilattice S. The order Id S is a join-semilattice and the same join formula holds as the one for lattices. Since the intersection of two ideals could be empty, Id S is not a lattice, in general. However, for a {∨, 0}-semilattice (a join-semilattice with zero), Id S is a lattice. For (join-semi-) lattices S and T , let ε : S → T be an embedding. We call ε an ideal-embedding if Sε is an ideal of T . Then, of course, for any ideal I of S, we have that Iε is an ideal of T . Ideal-embeddings play a major role in Chapter 18. By dualizing, we get the concepts of ﬁlter , fil(H), the ﬁlter generated by a subset H of the lattice L, provided that H = ∅, principal ﬁlter fil(a), and so on. 1.3.3

Congruences

An equivalence relation Θ on a lattice L is called a congruence relation, or congruence, of L iff a ≡ b (Θ) and c ≡ d (Θ) imply that (SP∨ ) (SP∧ )

a∧c≡b∧d

(Θ),

a∨c≡b∨d

(Θ)

(Substitution Properties). Trivial examples are ω and ι (introduced in Section 1.1.4). As in Section 1.1.4, for a ∈ L, we write a/Θ for the congruence class containing a; observe that a/Θ is a convex sublattice. If L is a lattice, K ≤ L, and Θ a congruence on L, then ΘK, the restriction of Θ to K, is a congruence of K. Formally, for x, y ∈ K, x ≡ y (ΘK)

iff x ≡ y (Θ) in L.

1.3. Some algebraic concepts

15

We call Θ discrete on K if ΘK = ω. Sometimes it is tedious to compute that a binary relation is a congruence relation. Such computations are often facilitated by the following lemma (Gr¨ atzer and E. T. Schmidt [56] and Maeda [96]): Lemma 1.1. A reﬂexive binary relation Θ on a lattice L is a congruence relation iff the following three properties are satisﬁed, for x, y, z, t ∈ L: (i) x ≡ y (Θ)

iff

x ∧ y ≡ x ∨ y (Θ).

(ii) x ≤ y ≤ z, x ≡ y (Θ), and y ≡ z (Θ) imply that x ≡ z (Θ). (iii) x ≤ y and x ≡ y (Θ) imply that x ∧ t ≡ y ∧ t (Θ) and x ∨ t ≡ y ∨ t (Θ). Let Con L denote the set of all congruence relations on L ordered by set inclusion (remember that we can view Θ ∈ Con L as a subset of L2 ). Theorem 1.2. Con L is a lattice. For Θ, Φ ∈ Con L, Θ ∧ Φ = Θ ∩ Φ. The join, Θ ∨ Φ, can be described as follows: x ≡ y (Θ ∨ Φ) iff there is a sequence z0 = x ∧ y ≤ z1 ≤ · · · ≤ zn = x ∨ y of elements of L such that zi ≡ zi+1 (Θ) or zi ≡ zi+1 (Φ), for every i with 0 ≤ i < n. Remark. For the binary relations α and β on a set A, we deﬁne the binary relation α ◦ β, the product of α and β, as follows: for a, b ∈ A, the relation a α ◦ β b holds iff a α x and x β b, for some x ∈ A. The relation Θ ∨ Φ is formed by repeated products. Theorem 1.2 strengthens this statement. The integer n in Theorem 1.2 can be restricted for some congruence joins. We call the congruences Θ and Φ permutable if Θ ∨ Φ = Θ ◦ Φ. A lattice L is congruence permutable if any pair of congruences of L are permutable. The chain Cn is congruence permutable iff n ≤ 2. Con L is called the congruence lattice of L. Observe that Con L is a sublattice of Part L; that is, the join and meet of congruence relations as congruence relations and as equivalence relations (partitions) coincide. If L is nontrivial, then Con L contains the two-element sublattice {ω, ι}. If Con L = {ω, ι}, we call the lattice L simple. All the nontrivial lattices of Figure 1.5 are simple. Of the many lattices of Figure 1.6, only M3 is simple. Given a, b ∈ L, there is a smallest congruence con(a, b)—called a principal congruence—under which a ≡ b. The formula (3) Θ = ( con(a, b) | a ≡ b (Θ) )

16

1. Basic Concepts

is trivial but important. For H ⊆ L, the smallest congruence under which H is in one class is con(H) = ( con(a, b) | a, b ∈ H ). Homomorphisms and congruence relations express two sides of the same phenomenon. Let L be a lattice and let Θ be a congruence relation on L. Let L/Θ = { a/Θ | a ∈ L }. Deﬁne ∧ and ∨ on L/Θ by a/Θ ∧ b/Θ = (a ∧ b)/Θ and a/Θ ∨ b/Θ = (a ∨ b)/Θ. The lattice axioms are easily veriﬁed. The lattice L/Θ is the quotient lattice of L modulo Θ. Lemma 1.3. The map ϕΘ : x → x/Θ,

for x ∈ L,

is a homomorphism of L onto L/Θ. The lattice K is a homomorphic image of the lattice L iff there is a homomorphism of L onto K. Theorem 1.4 (illustrated in Figure 1.8) states that any quotient lattice is a homomorphic image. To state it, we need one more concept: Let ϕ : L → L1 be a homomorphism of the lattice L into the lattice L1 , and deﬁne the binary relation Θ on L by x Θ y iff xϕ = yϕ; the relation Θ is a congruence relation of L, called the kernel of ϕ, in notation, ker(ϕ) = Θ. Theorem 1.4 (Homomorphism Theorem). Let L be a lattice. Any homomorphic image of L is isomorphic to a suitable quotient lattice of L. In fact, if ϕ : L → L1 is a homomorphism of L onto L1 and Θ is the kernel of ϕ, then L/Θ ∼ = L1 ; an isomorphism (see Figure 1.8 ) is given by ψ : x/Θ → xϕ, for x ∈ L.

ϕ L x

L1

onto

→ x/Θ L/Θ

x ψ:

/Θ

→

xϕ

Figure 1.8: The Homomorphism Theorem. We also know the congruence lattice of a homomorphic image: Theorem 1.5 (Second Isomorphism Theorem). Let L be a lattice and let Θ be a congruence relation of L. For any congruence Φ of L such that Φ ≥ Θ, deﬁne the relation Φ/Θ on L/Θ by x/Θ ≡ y/Θ

(Φ/Θ)

iff

x≡y

(Φ).

1.3. Some algebraic concepts

17

Then Φ/Θ is a congruence of L/Θ. Conversely, every congruence Ψ of L/Θ can be (uniquely) represented in the form Ψ = Φ/Θ, for some congruence Φ ≥ Θ of L. In particular, the congruence lattice of L/Θ is isomorphic with the interval [Θ, ι] of the congruence lattice of L. Let L be a bounded lattice. A congruence Θ of L separates 0 if 0/Θ = {0}, that is, x ≡ 0 (Θ) implies that x = 0. Similarly, a congruence Θ of L separates 1 if 1/Θ = {1}, that is, x ≡ 1 (Θ) implies that x = 1. We call the lattice L non-separating if 0 and 1 are not separated by any congruence Θ = ω. Similarly, a homomorphism ϕ of the lattices L1 and L2 with zero is 0separating if 0ϕ = 0, but xϕ = 0, for x = 0.

2 Special Concepts

In this chapter we introduce special elements, constructions, and classes of lattices that play an important role in the representation of ﬁnite distributive lattices as congruence lattices of ﬁnite lattices.

2.1.

Elements and lattices

In a nontrivial ﬁnite lattice L, an element a is join-reducible if a = 0 or if a = b ∨ c for some b < a and c < a; otherwise, it is join-irreducible. Let J(L) denote the set of all join-irreducible elements of L, regarded as an order under the ordering of L. For a ∈ L, set J(a) = { x | x ≤ a, x ∈ J(L) } = id(a) ∩ J(L), that is, J(a) is ↓ a formed in J(L). Note that, by deﬁnition, 0 is not a joinirreducible element; and similarly, 1 is not a meet-irreducible element. In a ﬁnite lattice, every element is a join of join-irreducible elements (indeed, a = J(a)), and similarly for meets. Dually, we deﬁne meet-reducible and meet-irreducible elements. An element a is an atom if 0 ≺ a and a dual atom if a ≺ 1. Atoms are join-irreducible. A lattice L is atomistic if every element is a ﬁnite join of atoms. For inﬁnite lattices, we have a related concept: A lattice L is weakly atomic if every nontrivial interval contains a prime interval. In a bounded lattice L, the element a is a complement of the element b iff a ∧ b = 0 and a ∨ b = 1. A complemented lattice is a bounded lattice in

20

2. Special Concepts

Figure 2.1: The lattice N6 . which every element has a complement; the lattices of Figure 1.5 are complemented and so are all but one of the lattices of Figure 1.6. The lattice Bn is complemented. Let a ∈ [b, c]; the element x is a relative complement of a in [b, c] iff a ∧ x = b, and a ∨ x = c. A relatively complemented lattice is a lattice in which every element has a relative complement in any interval containing it. The lattice N5 is complemented but not relatively complemented. In a lattice L with zero, let a ≤ b. A complement of a in [0, b] is called a sectional complement of a in b. A lattice L with zero is called sectionally complemented if a has a sectional complement in b, for all a ≤ b in L. The lattice N6 of Figure 2.1 is sectionally complemented but not relatively complemented.

2.2.

Direct and subdirect products

Let L and K be lattices and form the direct product L × K as in Section 1.1.3. Then L × K is a lattice and ∨ and ∧ are computed “componentwise”: a, b ∨ c, d = a ∨ c, b ∨ d, a, b ∧ c, d = a ∧ c, b ∧ d. See Figure 2.2 for the example C2 × N5 . Obviously, Bn is isomorphic to a direct product of n copies of B1 . There are two projection maps (homomorphisms) associated with this construction: πL : L × K → L and πK : L × K → K, deﬁned by πL : x, y → x and by πK : x, y → y, respectively. Similarly, we can form the direct product L1 × · · · × Ln with elements x1 , . . . , xn , where xi ∈ Li , for i ≤ n; we denote the projection map x1 , . . . , xi , . . . , xn → xi

2.2. Direct and subdirect products

21

N5 C2

Figure 2.2: C2 × N5 , a direct product of two lattices. by πi . If Li = L, for all i ≤ n, we get the direct power Ln . By identifying x ∈ Li with 0, . . . , 0, x, 0, . . . , 0 (x is the i-th coordinate), we regard Li as an ideal of L1 × · · · × Ln , for i ≤ n. The black-ﬁlled elements in Figure 2.2 show how we consider C2 and N5 ideals of C2 × N5 . A very important property of direct products is: Theorem 2.1. Let L and K be lattices, let Θ be a congruence relation of L, and let Φ be a congruence relation of K. Deﬁne the relation Θ × Φ on L × K by a1 , b1 ≡ a2 , b2 (Θ × Φ) iff a1 ≡ a2 (Θ) and b1 ≡ b2 (Φ). Then Θ × Φ is a congruence relation on L × K. Conversely, every congruence relation of L × K is of this form. A more general construction is subdirect products. If L ≤ K1 × · · · × Kn and the projection maps πi are onto maps, for i ≤ n, then we call L a subdirect product of K1 , . . . , Kn . Trivial examples: L is a subdirect product of L and L if we identify x ∈ L with x, x ∈ L2 (diagonal embedding). In this example, the projection map is an isomorphism. To exclude such trivial cases, let us call a representation of L as a subdirect product of K1 , . . . , Kn trivial if one of the projection maps πi is an isomorphism, for i ≤ n. A lattice L is called subdirectly irreducible iff all representations of L as a subdirect product are trivial. Let L be a subdirect product of K1 and K2 ; then ker(π1 ) ∧ ker(π2 ) = ω. This subdirect product is trivial iff ker(π1 ) = ω or ker(π2 ) = ω. Conversely, if Φ1 ∧ Φ2 = ω in Con A, then K is a subdirect product of K/Φ1 and K/Φ2 , and this representation is trivial iff Φ1 = ω or Φ2 = ω.

22

2. Special Concepts

1

ι

L

Con L

v u

0

con(u, v)

ω

Figure 2.3: A subdirectly irreducible lattice and its congruence lattice. Every simple lattice is subdirectly irreducible. The lattice N5 is subdirectly irreducible but not simple. There is a natural correspondence between subdirect representations of a lattice L and a set of congruences {Φ1 , . . . , Φn } satisfying Φ1 ∧ · · · ∧ Φn = ω. This representation is nontrivial iff Φi = ω, for all i ≤ n. In this subdirect representation, the factors (the lattices L/Φi ) are subdirectly irreducible iff the congruences Φi are meet-irreducible, for all i ≤ n, by the Second Isomorphism Theorem (Theorem 1.5). For a ﬁnite lattice L, the lattice Con L is ﬁnite, so we can represent ω as a meet of meet-irreducible congruences, and we obtain: Theorem 2.2. Every ﬁnite lattice L is a subdirect product of subdirectly irreducible lattices. This result (Birkhoﬀ’s Subdirect Representation Theorem) is true for any algebra in any variety (a class of algebras deﬁned by identities, such as the class of all lattices). Finite subdirectly irreducible lattices are easy to recognize. If L is such a lattice, then the meet Θ of all the > ω elements is > ω. Obviously, Θ is an atom, the unique atom of Con L. Conversely, if Con L has a unique atom, then all > ω congruences are ≥ Θ, so their meet cannot be ω. We call Θ the base congruence of L (called monolith in many publications). If u = v and u ≡ v (Θ), then Θ = con(u, v). So Con L = {ω} ∪ [con(u, v), ι], as illustrated in Figure 2.3. Let L be a ﬁnite subdirectly irreducible lattice with con(u, v) the base congruence, where u ≺ v ∈ L. By adding two elements as shown in Figure 2.4, we embed L into a simple lattice.

2.3. Polynomials and identities

23

1

L a

v u b 0

Figure 2.4: Embedding into a simple lattice. Lemma 2.3. Every ﬁnite subdirectly irreducible lattice can be embedded into a simple lattice with at most two extra elements. Note that every ﬁnite lattice can be embedded into a ﬁnite simple lattice; in general, we need more than two elements. For a stronger statement, see Lemma 12.3.

2.3.

Polynomials and identities

From the variables x1 , . . . , xn , we can form (n-ary) polynomials (also called terms or polynomials without parameters) in the usual manner using ∨, ∧, and parentheses. Examples of polynomials are: x1 , x3 , x1 ∨x1 , (x1 ∧x2 )∨(x3 ∧x1 ), (x3 ∧ x1 ) ∨ ((x3 ∨ x2 ) ∧ (x1 ∨ x2 )). An n-ary polynomial p deﬁnes a function in n variables (a polynomial function, or simply, a polynomial ) on a lattice L. For example, if p = (x1 ∧ x3 ) ∨ (x3 ∨ x2 ) and a, b, c ∈ L, then p(a, b, c) = (a ∧ c) ∨ (c ∨ b) = b ∨ c. If p is a unary (n = 1) lattice polynomial, then p(a) = a, for any a ∈ L. If p is binary, then p(a, b) = a, or p(a, b) = b, or p(a, b) = a ∧ b, or p(a, b) = a ∨ b, for all a, b ∈ L. If p = p(x1 , . . . , xn ) is an n-ary polynomial and L is a lattice, then by substituting some variables by elements of L, we get a function on L of nvariables. Such functions are called algebraic functions. Unary algebraic functions of the form p(x) = p(a1 , . . . , ai−1 , x, ai+1 , . . . , xn ),

24

2. Special Concepts

where a1 , . . . , ai−1 , ai+1 , . . . , an ∈ L, play the most important role, see Section 3.1. A polynomial (function), in fact, any algebraic function, p is isotone; that is, if a1 ≤ b1 , . . . , an ≤ bn , then p(a1 , . . . , an ) ≤ p(b1 , . . . , bn ). Furthermore, a1 ∧ · · · ∧ an ≤ p(a1 , . . . , an ) ≤ a1 ∨ · · · ∨ an . Polynomials have many uses. We brieﬂy discuss three. (i) The sublattice generated by a set Lemma 2.4. Let L be a lattice and let H be a nonempty subset of L. Then a ∈ sub(H) (the sublattice generated by H) iff a = p(h1 , . . . , hn ), for some integer n ≥ 1, for some n-ary polynomial p, and for some h1 , . . . , hn ∈ H. (ii) Identities A lattice identity (resp., lattice inequality)—also called equation—is an expression of the form p = q (resp., p ≤ q), where p and q are polynomials. An identity p = q (resp., p ≤ q) holds in the lattice L iff p(a1 , . . . , an ) = q(a1 , . . . , an ) (resp., p(a1 , . . . , an ) ≤ q(a1 , . . . , an )) holds, for all a1 , . . . , an ∈ L. The identity p = q is equivalent to the two inequalities p ≤ q and q ≤ p; the inequality p ≤ q is equivalent to the identity p ∨ q = q. The most important properties of identities are given by Lemma 2.5. Identities are preserved under the formation of sublattices, homomorphic images, direct products, and ideal lattices. A lattice L is called distributive if the identities x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z), x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) hold in L. In fact, it is enough to assume one of these identities, because the two identities are equivalent. As we have just noted, the identity x∧(y∨z) = (x∧y)∨(x∧z) is equivalent to the two inequalities: x ∧ (y ∨ z) ≤ (x ∧ y) ∨ (x ∧ z), (x ∧ y) ∨ (x ∧ z) ≤ x ∧ (y ∨ z). However, the second inequality holds in any lattice. So a lattice is distributive iff the inequality x ∧ (y ∨ z) ≤ (x ∧ y) ∨ (x ∧ z) holds. By duality, we get a similar statement about the second identity deﬁning distributivity. The class of all distributive lattices will be denoted by D. A boolean lattice is a distributive complemented lattice. A ﬁnite boolean lattice is isomorphic to some Bn .

2.3. Polynomials and identities

25

A lattice is called modular if the identity (x ∧ y) ∨ (x ∧ z) = x ∧ (y ∨ (x ∧ z)) holds. Note that this identity is equivalent to the following implication: x ≥ z implies that (x ∧ y) ∨ z = x ∧ (y ∨ z). The class of all modular lattices will be denoted by M. Every distributive lattice is modular. The lattice M3 is modular but not distributive. All the lattices of Figures 1.5 and 1.6 are modular except for Part {1, 2, 3, 4} and N5 . A class of lattices V is called a variety if it is deﬁned by a set of identities. The classes D and M are examples of varieties, and so are L, the variety of all lattices and T, the (trivial) variety of one-element lattices. (iii) Free lattices Starting with a set H, we can form the set of all polynomials over H, collapsing two polynomials if their equality follows from the lattice axioms. We thus form the “free-est” lattice over H. For instance, if we start with H = {a, b}, then we obtain the four-element lattice, F (2), of Figure 2.5. We obtain more interesting examples if we start with an order P and require that the ordering in P be preserved. For instance, if we start with P = {a, b, c} with a < b, then we get the corresponding nine-element “free” lattice, F (P ), of Figure 2.5.

b

c

a

F (2)

F (P )

Figure 2.5: Two free lattices. Sometimes, we need free lattices with respect to some special conditions. The following result illustrates this. Lemma 2.6. Let x, y, and z be elements of a lattice L and let x ∨ y, y ∨ z, z ∨ x be pairwise incomparable. Then sub({x ∨ y, y ∨ z, z ∨ x}) ∼ = B3 .

26

2. Special Concepts

Lemma 2.6 is illustrated by the following diagram:

x∨y∨z

x∨y

y∨z

z∨x

(y ∨ z) ∧ (z ∨ x)

(x ∨ y) ∧ (y ∨ z) (x ∨ y) ∧ (z ∨ x)

(x ∨ y) ∧ (y ∨ z) ∧ (z ∨ x) Figure 2.6: A free lattice with special relations. We will also need “free distributive lattices”, obtained by collapsing two polynomials if their equality follows from the lattice axioms and the distributive identities. Starting with a three-element set H = {x, y, z}, we then obtain the lattice, FD (3), of Figure 2.7. Similarly, we can deﬁne “free modular lattices.” Starting with a threeelement set H = {x, y, z}, we then obtain the lattice, FM (3), of Figure 2.8. An equivalent deﬁnition of freeness is the following: Let H be a set and let K be a variety of lattices. A lattice FK (H) is called a free lattice over K generated by H iff the following three conditions are satisﬁed: (i) FK (H) ∈ K. (ii) FK (H) is generated by H. (iii) Let L ∈ K and let ϕ : H → L be a map; then there exists a (lattice) homomorphism ψ : FK (H) → L extending ϕ (that is, satisfying aϕ = aψ, for all a ∈ H).

2.4.

Gluing

In Section 1.1.3 glued sums of orders, P Q, applied to an order P with largest element 1P and an order Q with smallest element 0Q were introduced. This applies to any two lattices K with a unit and L with a zero. A natural generalization of this construction is gluing.

2.4. Gluing

27

a = (x ∨ y) ∧ (x ∨ z) ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z) ∨ (y ∧ z) x∨y∨z

b = (x ∨ y) ∧ (y ∨ z) c = (x ∧ y) ∨ (y ∧ z)

x∨y

x∨z b

(x ∨ y) ∧ (x ∨ z)

a

x

(x ∧ y) ∨ (x ∧ z) x∧y

y∨z

(x ∨ z) ∧ (y ∨ z)

y

c x∧z

z

(x ∧ z) ∨ (y ∧ z) y∧z

x∧y∧z Figure 2.7: The free distributive lattice on three generators, FD (3). Let K and L be lattices, let F be a ﬁlter of K, and let I be an ideal of L. If F is isomorphic to I (with ψ the isomorphism), then we can form the lattice G, the gluing of K and L over F and I (with respect to ψ), deﬁned as follows: We form the disjoint union K ∪ L, and identify a ∈ F with aψ ∈ I, for all a ∈ F , to obtain the set G. We order G as follows (see Figure 2.9): ⎧ if a, b ∈ K; a ≤K b, ⎪ ⎪ ⎪ ⎨a ≤ b, if a, b ∈ L; L a ≤ b iff ⎪ a ≤K x and xψ ≤L b, if a ∈ K and b ∈ L, ⎪ ⎪ ⎩ for some x ∈ F .

28

2. Special Concepts

u = (x ∧ y) ∨ (y ∧ z) ∨ (x ∧ z) v = (x ∨ y) ∧ (y ∨ z) ∧ (x ∨ z) x1 = (x ∧ v) ∨ u y1 = (y ∧ v) ∨ u z1 = (z ∧ v) ∨ u

x∨u

x

y∨u

v

x1 x∧v

y1

u

z∨u

y

z1 y∧v

z

z∧v

Figure 2.8: The free modular lattice on three generators, FM (3).

Lemma 2.7. G is an order, in fact, G is a lattice. The join in G is described by ⎧ ⎪ ⎨a ∨K b, a ∨G b = a ∨L b, ⎪ ⎩ (a ∨K x)ψ ∨L b,

if a, b ∈ K; if a, b ∈ L; if a ∈ K and b ∈ L, for any b ≥ x ∈ F ,

and dually for the meet. If L has a zero, 0L , then the last clause for the join

2.4. Gluing

b

29

(a ∨K x) ∨L b b

L

L

y

I

I x

x

a

K

K

a

y = a ∨K x

G

Figure 2.9: Deﬁning gluing.

I

L

F

G K Figure 2.10: An easy gluing example. may be rephrased: a ∨G b = (a ∨K 0L )ψ ∨L b,

if a ∈ K and b ∈ L.

G contains K and L as sublattices; in fact, K is an ideal and L is a ﬁlter of G. An example of gluing is shown in Figure 2.10. There are more sophisticated examples in this book; see, for instance, Chapters 10 and 16. Lemma 2.8. Let K, L, F , I, and G be given as above. Let A be a lattice containing K and L as sublattices so that K ∩ L = I = F . Then K ∪ L is a sublattice of A and it is isomorphic to G.

30

2. Special Concepts

Now if ΘK is a binary relation on K and ΘL is a binary relation on L, we r deﬁne the reﬂexive product ΘK ◦ ΘL as ΘK ∪ ΘL ∪ (ΘK ◦ ΘL ). We can easily describe the congruences of G. Lemma 2.9. A congruence Θ of G can be uniquely written in the form r

Θ = ΘK ◦ ΘL , where ΘK is a congruence of K and ΘL is a congruence of L satisfying the condition that ΘK restricted to F equals ΘL restricted to I (under the identiﬁcation of elements by ψ). Conversely, if ΘK is a congruence of K and ΘL is a congruence of L satisfying the condition that ΘK restricted to F equals ΘL restricted to I, r then Θ = ΘK ◦ ΘL is a congruence of G. Let A and B be lattices, FA a ﬁlter of A, IB an ideal of B, and FB a ﬁlter of B. Let us assume that FA , IB , and FB are isomorphic. We now deﬁne what it means to obtain C by gluing B to A k-times. For k = 1, let C be the gluing of A and B over FA and IB with the ﬁlter FB regarded as a ﬁlter FC of C. Now if Ck−1 with the ﬁlter FCk−1 is the gluing of B to A k − 1-times, then we glue Ck−1 and B over FCk−1 and IB to obtain C the gluing of B to A k-times with the ﬁlter FB regarded as a ﬁlter FC of C.

2.5.

Modular and distributive lattices

2.5.1

The characterization theorems

The two typical examples of nondistributive lattices are N5 and M3 , whose diagrams are given (again) in Figure 2.11. The following characterization theorem follows immediately by inspecting the diagrams of the free lattices in Figures 2.5 and 2.8.

i i

b c a

a

N5

o

c

b o

M3

Figure 2.11: The two characteristic nondistributive lattices.

2.5. Modular and distributive lattices

31

Theorem 2.10. (i) A lattice L is modular iff it does not contain N5 as a sublattice. (ii) A modular lattice L is distributive iff it does not contain M3 as a sublattice. (iii) A lattice L is distributive iff L contains neither N5 nor M3 as a sublattice. Theorem 2.11. Let L be a modular lattice, let a ∈ L, and let U and V be sublattices with the property u ∧ v = a, for all u ∈ U and v ∈ V . Then sub(U ∪ V ) is isomorphic to U × V under the isomorphism (u ∈ U and v ∈ V ) u ∨ v → u, v. Conversely, a lattice L satisfying this property is modular. Corollary 2.12. Let L be a modular lattice and let a, b ∈ L. Then sub([a ∧ b, a] ∪ [a ∧ b, b]), that is, the sublattice of L generated by [a ∧ b, a] ∪ [a ∧ b, b], is isomorphic to the direct product [a ∧ b, a] × [a ∧ b, b]. In the distributive case, the sublattice generated by [a ∧ b, a] ∪ [a ∧ b, b] is the interval [a ∧ b, a ∨ b]; this does not hold for modular lattices, as exempliﬁed by M3 . Let G be the gluing of the lattices K and L over F and I, as in Section 2.4. Lemma 2.13. If K and L are modular, so is the gluing G of K and L. If K and L are distributive, so is G. The distributive identity easily implies that every n-ary polynomial equals one we get by joining meets of variables. So we get: Lemma 2.14. A ﬁnitely generated distributive lattice is ﬁnite. 2.5.2

Finite distributive lattices

For a nontrivial ﬁnite distributive lattice D, the order J(D) is “equivalent” to D in the following sense: Theorem 2.15. Let D be a nontrivial ﬁnite distributive lattice. Then the map ψ : a → J(a) is an isomorphism between D and Down(J(D)).

32

2. Special Concepts

Corollary 2.16. The correspondence D → J(D) makes the class of all nontrivial ﬁnite distributive lattices correspond to the class of all ﬁnite orders; isomorphic lattices correspond to isomorphic orders, and vice versa. In particular, D ∼ = J(Down P ). = Down(J(D)) and P ∼ Let D and E be nontrivial ﬁnite distributive lattices, and let ϕ : D → E be a {0, 1}-homomorphism. Then with every x ∈ J(E), we can associate the smallest y ∈ D with yϕ ≥ x. It turns out that y ∈ J(D), so we obtain an isotone map J(ϕ) : J(E) → J(D). Theorem 2.17. Let D and E be nontrivial ﬁnite distributive lattices, and let ϕ : D → E be a {0, 1}-homomorphism. Let ϕD and ϕE be the isomorphisms between D and Down(J(D)) and between E and Down(J(E)), respectively. Then the diagram ψD D −−−−→ Down(J(D)) ⏐ ⏐ ⏐ ⏐Down(J(ϕ)) ϕ ψE

E −−−−→ Down(J(E)) commutes, that is, ψD Down(J(ϕ)) = ϕψE . Let U be a ﬁnite order. If U is an antichain, then Pow U ∼ = Down U , the ﬁnite boolean lattice of the power set of H. Since Down(J(D)) is a sublattice of the ﬁnite boolean lattice of all subsets of J(D), we get Corollary 2.18. Every ﬁnite distributive lattice D can be embedded into a ﬁnite boolean lattice. If D is nontrivial, then it can be embedded into Bn , where n = |J(D)| In Theorem 2.15, instead of J(D) and down-sets, we could work with the dual concepts: M(D) (the order of meet-irreducible elements of D) and upsets. However, surprisingly, J(D) and M(D) are isomorphic. To see this, for a ∈ M(D), let a† denote the smallest element x of D not below a (that is, with x a). By distributivity, it is easy to see that a† ∈ J(D); in fact, we have the following. Theorem 2.19. Let D be a nontrivial ﬁnite distributive lattice. Then the map ψ : a → a† is an isomorphism between the orders M(D) and J(D). 2.5.3

Finite modular lattices

Take a look at the two positions of the pair of intervals [a, b] and [c, d] in Figure 2.12. In either case, we will write [a, b] ∼ [c, d], and say that [a, b] is perspective to [c, d]. If we want to show whether the perspectivity is “up” or

2.5. Modular and distributive lattices

d=b ∨c

b

c

33

b=a∨d

a

d

b∧c=a

c=a∧d u

d

Figure 2.12: [a, b] ∼ [c, d] and [a, b] ∼ [c, d]. u

d

“down”, we will write [a, b] ∼ [c, d] in the ﬁrst case and [a, b] ∼ [c, d] in the second case. If, for some natural number n and intervals [ei , fi ], for 0 ≤ i ≤ n, [a, b] = [e0 , f0 ] ∼ [e1 , f1 ] ∼ · · · ∼ [en , fn ] = [c, d], then we say that [a, b] is projective to [c, d] and write [a, b] ≈ [c, d]. One of the most important properties of a modular lattice is stated in the following result: Theorem 2.20 (Isomorphism Theorem for Modular Lattices). Let L u be a modular lattice and let [a, b] ∼ [c, d] in L. Then ϕc : x → x ∨ c,

x ∈ [a, b],

is an isomorphism of [a, b] and [c, d]. The inverse isomorphism is ψb : y → y ∧ b,

y ∈ [c, d].

(See Figure 2.13.) Corollary 2.21. In a modular lattice, projective intervals are isomorphic. Corollary 2.22. In a modular lattice if a prime interval p is projective to an interval q, then q is also prime. Let us call the ﬁnite lattice L semimodular or upper semimodular if for a, b, c ∈ L, the covering a ≺ b implies that a ∨ c ≺ b ∨ c or a ∨ c = b ∨ c. The dual of an upper semimodular lattice is a lower semimodular lattice. Lemma 2.23. A modular lattice is both upper and lower semimodular. For a ﬁnite lattice, the converse also holds: a ﬁnite upper and lower semimodular lattice is modular, and conversely. The lattice S8 , in Figure 9.1 (see page 105), is an example of a semimodular lattice that is not modular. See Section 9.2 for an interesting use of this lattice. The following is even more trivial than Lemma 2.13:

34

2. Special Concepts

d y x∨c ψb

b y∧b

c ϕc

x a Figure 2.13: The isomorphisms ϕc and ψb . Lemma 2.24. If K and L are ﬁnite semimodular lattices, so is the gluing G of K and L. A large class of semimodular but not modular lattices is provided by Lemma 2.25. Let A be a nonempty set. Then Part A is semimodular; it is not modular unless |A| ≤ 3.

Part II

Basic Techniques

3 Congruences

3.1.

Congruence spreading

Let a, b, c, and d be elements of a lattice L. If a ≡ b (Θ)

implies that c ≡ d (Θ),

for any congruence relation Θ of L, then we can say that a ≡ b congruenceforces c ≡ d. In Section 1.3.3 we saw that a ≡ b (Θ) iff a ∧ b ≡ a ∨ b (Θ); therefore, to investigate congruence-forcing, it is enough to deal with comparable pairs, a ≤ b and c ≤ d. Instead of comparable pairs, we will deal with intervals [a, b] and [c, d]. Projectivity (see Section 2.5.3) is suﬃcient for the study of congruenceforcing (or congruence spreading) in some classes of lattices (for instance, in the class of modular lattices). In general, however, we have to introduce somewhat more general concepts and notation. As illustrated in Figure 3.1, we say that [a, b] is up congruence-perspective u onto [c, d] and write [a, b] [c, d] iff there is an a1 ∈ [a, b] with [a1 , b] ∼ [c, d]; similarly, [a, b] is down congruence-perspective onto [c, d] and write d

[a, b] [c, d] iff there is a b1 ∈ [a, b] with [a, b1 ] ∼ [c, d]. If [a, b] [c, d] or [a, b] [c, d], then [a, b] is congruence-perspective onto [c, d] and we write [a, b] → [c, d]. If, for some natural number n and and intervals [ei , fi ], for 0 ≤ i ≤ n, [a, b] = [e0 , f0 ] → [e1 , f1 ] → · · · → [en , fn ] = [c, d],

36

3. Congruences

b

d=b∨c

b

c

b1 = a ∨ d

a1 = b ∧ c

a

d c =a∧d

a

Figure 3.1: [a, b] [c, d] and [a, b] [c, d] then we call [a, b] congruence-projective onto [c, d], and we write [a, b] ⇒ [c, d]. If [a, b] ⇒ [c, d] and [c, d] ⇒ [a, b], then we write [a, b] ⇔ [c, d]. u Also if (see Section 2.5.3) [a, b] ∼ [c, d], then [a, b] [c, d] and [c, d] [a, b]; d

if [a, b] ∼ [c, d], then [a, b] [c, d] and [c, d] [a, b]; if [a, b] ∼ [c, d], then [a, b] → [c, d] and [c, d] → [a, b]; if [a, b] ≈ [c, d], then [a, b] ⇒ [c, d] and [c, d] ⇒ [a, b]. But while ∼, ≈, and ⇔ are symmetric, the relations → and ⇒ are not. In particular, if a ≤ c ≤ d ≤ b, then [a, b] ⇒ [c, d]. If [a, b] ⇒ [c, d], then there is a unary algebraic function p with p(a) = c and p(c) = d. It is easy to see what special kinds of unary algebraic functions are utilized in the relation ⇒. Intuitively, “a ≡ b congruence-forces c ≡ d ” iff [c, d] is put together from pieces [c , d ] each satisfying [a, b] ⇒ [c , d ]. To state this more precisely, we describe con(a, b), the smallest congruence relation under which a ≡ b (introduced in Section 1.3.3); see Dilworth [12]. Theorem 3.1. Let L be a lattice, a, b, c, d ∈ L, a ≤ b, c ≤ d. Then c ≡ d (con(a, b)) iff, for some sequence c = e0 ≤ e1 ≤ · · · ≤ em = d, we have [a, b] ⇒ [ej , ej+1 ],

for

j = 0, . . . , m − 1.

Let L be a lattice and H ⊆ L . To compute con(H), the smallest congruence relation Θ under which a ≡ b (Θ), for all a, b ∈ H, we use the formula con(H) = ( con(a, b) | a, b ∈ H ). 2

We also need a formula for joins: Lemma 3.2. Let L be a lattice and let Θi , i ∈ I, be congruence relations of L. Then a ≡ b ( ( Θi | i ∈ I )) iff there is a sequence z0 = a ∧ b ≤ z1 ≤ · · · ≤ zn = a ∨ b

3.2. Prime intervals

37

such that, for each j with 0 ≤ j < n, there is an ij ∈ I satisfying zj ≡ zj+1 (Θij ). This is an easy but profoundly important result. For instance, the wellknown result of Funayama and Nakayama [21]—which provides the foundation for this book—immediately follows. (Another typical application is Lemma 3.10.) Theorem 3.3. The lattice Con L is distributive, for any lattice L. Proof. Let Θ, Φ, Ψ ∈ Con L. As we note on page 24, to verify the identity Θ ∧ (Φ ∨ Ψ) = (Θ ∧ Φ) ∨ (Θ ∧ Ψ), it is suﬃcient to verify the inequality Θ ∧ (Φ ∨ Ψ) ≤ (Θ ∧ Φ) ∨ (Θ ∧ Ψ). So let a ≡ b (Θ ∧ (Φ ∨ Ψ)). Then a ≡ b (Θ) and so a ∧ b ≡ a ∨ b (Θ), and also a ≡ b (Φ ∨ Ψ), so by Lemma 3.2, there is a sequence z0 = a ∧ b ≤ z1 ≤ · · · ≤ zn = a ∨ b such that zj ≡ zj+1 (Φ) or zj ≡ zj+1 (Ψ), for each j with 0 ≤ j < n. For each j, we have zj ≡ zj+1 (Θ), so either zj ≡ zj+1 (Θ ∧ Φ) or zj ≡ zj+1 (Θ ∧ Ψ), proving that a ≡ b ((Θ ∧ Φ) ∨ (Θ ∧ Ψ)). By combining Theorem 3.1 and Lemma 3.2, we get: Corollary 3.4. Let L be a lattice, let H ⊆ L2 , and let a, b ∈ L with a ≤ b. Then a ≡ b (con(H)) iff, for some integer n, there exists a sequence a = c0 ≤ c1 ≤ · · · ≤ cn = b such that, for each i with 0 ≤ i < n, there exists a di , ei ∈ H satisfying [di ∧ ei , di ∨ ei ] ⇒ [ci , ci+1 ].

3.2.

Prime intervals

Prime intervals play a dominant role in the study of the congruences of a ﬁnite lattice L. Let Prime(L) denote the set of prime intervals of L. Let p = [a, b] be a prime interval in L (that is, a ≺ b in L) and let Θ be a congruence relation of L; then we write p ∈ Θ, for a ≡ b (Θ). In a ﬁnite lattice L, the formula Θ = ( con(p) | p ∈ Θ ) immediately yields that the congruences in J(Con L) (to be denoted by ConJ L) are the congruences of the form con(p), for some p ∈ Prime(L). Of course, a join-irreducible congruence Θ can be expressed, as a rule, in many ways in the form con(p). However, Corollary 3.4 yields the following crucial statements:

38

3. Congruences

Lemma 3.5. Let L be a ﬁnite lattice, let p be a prime interval of L, and let [a, b] be an interval of L. If p is collapsed under con(a, b), then there is a prime interval q in [a, b] satisfying q ⇒ p. Lemma 3.6. Let L be a ﬁnite lattice, and let p and q be prime intervals of L. Then con(p) ≤ con(q) iff q ⇒ p and con(p) = con(q) iff p ⇔ q. In view of Theorem 2.15, we get the following: Theorem 3.7. Let L be a ﬁnite lattice. The relation ⇒ is a preordering on Prime(L). The equivalence classes under ⇔ form an order isomorphic to ConJ L. If the ﬁnite lattice L is atomistic, then the join-irreducible congruences are even simpler to ﬁnd. Indeed if [a, b] is a prime interval, and p is an atom d

with p ≤ b and p a, then [a, b] ∼ [0, p], so con(a, b) = con(0, p). Corollary 3.8. Let L be a ﬁnite atomistic lattice. Then every join-irreducible congruence can be represented in the form con(0, p), where p is an atom. The relation p ⇒ q, deﬁned as [0, p] ⇒ [0, q], introduces a preordering on the set of atoms of L. The equivalence classes under the preordering form an order isomorphic to ConJ L. We use Figure 3.2 to illustrate how we compute the congruence lattice of N5 using Theorem 3.7. N5 has ﬁve prime intervals: [o, a], [a, b], [b, i], [o, c], [c, i]. The equivalence classes are α = {[a, b]}, β = {[o, c], [b, i]}, and γ = {[o, a], [c, i]}. The ordering: α < γ because [c, i] [o, b] [a, b]. Similarly, α < β.

ι

i β

γ β

b α a

γ

c

γ

β o

N5

β

γ α ConJ N5

α ω Con N5

Figure 3.2: Computing the congruence lattice of N5 . It is important to note that the computation of p ⇒ q may involve nonprime intervals. For instance, let p = [o, c] and q = [a, b] in N5 . Then p ⇒ q, because p [a, i] q, but we cannot get p ⇒ q involving only prime intervals.

3.3. Congruence-preserving extensions and variants

39

β=ι α

β

α α β α α αα α β

S8

β

α

α

ω

ConJ S8

Con S8

Figure 3.3: Computing the congruence lattice of S8 . As another example, we compute the congruence lattice of S8 ; see Figure 3.3. By Corollary 2.23, in a modular lattice if p and q are prime intervals, then u d p q implies that p ∼ q and p q implies that p ∼ q; thus p ⇒ q implies that p ≈ q. Therefore, Theorem 3.7 tells us that ConJ L is an antichain in a ﬁnite modular lattice L, so Con L is boolean (as noted in Section 2.5.2). Corollary 3.9. The congruence lattice of a ﬁnite modular lattice is boolean. By a colored lattice we will mean a ﬁnite lattice (some) of whose prime intervals are labeled so that if the prime intervals p and q are of the same color, then con(p) = con(q). These labels represent (a subset of) the equivalence classes of prime intervals, as stated in Theorem 3.7. In Figure 3.3, every prime interval of S8 is labeled; in Figure 8.2, only some are labeled. The coloring helps in the intuitive understanding of some constructions. If the prime intervals p is of color c, then we deﬁne con(c) as con(p).

3.3.

Congruence-preserving extensions and variants

Let L be a lattice, and K ≤ L. How do the congruences of L relate to the congruences of K? Every congruence Θ reﬂects (or restricts) to K: the relation Θ∩K 2 = ΘK on K is a congruence of K. So we get the reﬂection map (also called restriction map): re : Con L → Con K, that maps a congruence Θ of L to ΘK. Lemma 3.10. Let K ≤ L be lattices. Then re : Con L → Con K is a {∧, 0, 1}homomorphism. For instance, if K = {o, a, i} and L = M3 (see Figure 2.11), then Con K is isomorphic to B2 , but only ω and ι are reﬂections of congruences in L. As another example, take the lattice L of Figure 3.4 and its sublattice K, the black-ﬁlled elements; in this case, Con L ∼ = B2 , but again only ω and = Con K ∼

40

3. Congruences

ι are reﬂections. There is no natural relationship between the congruences of K and L. If K is an ideal in L (or any convex sublattice), we can say a lot more.

K

L

Figure 3.4: Illustrating the map re. Lemma 3.11. Let K ≤ L be lattices. If K is an ideal of L, then re : Con L → Con K is a {0, 1}-homomorphism. Proof. By Lemma 3.10, the map re is a {∧, 0, 1}-homomorphism. Let Θ and Φ be congruences of L; we have to prove that ΘK ∨ ΦK = (Θ ∨ Φ)K. Since ≤ is trivial, we prove ≥. So let a, b ∈ K, a ≡ b ((Θ ∨ Φ)K); we want to prove that a ≡ b (ΘK ∨ ΦK). By Lemma 3.2, there is a sequence z 0 = a ∧ b ≤ z1 ≤ · · · ≤ z n = a ∨ b such that, for each j with 0 ≤ j < n, either zj ≡ zj+1 (Θ) or zj ≡ zj+1 (Φ) holds in L. Since a, b ∈ K and K is an ideal, it follows that z0 , z1 , . . . , zn ∈ K, so for each j with 0 ≤ j < n, either zj ≡ zj+1 (ΘK) or zj ≡ zj+1 (ΦK) holds, proving that a ≡ b (ΘK ∨ ΦK). Let K ≤ L be lattices, and let Θ be a congruence of K. The congruence conL (Θ) (the congruence con(Θ) formed in L) is the smallest congruence Φ of L such that Θ ≤ ΦK. Unfortunately, conL (Θ)K may be diﬀerent from Θ, as in the example of Figure 3.4. We say that the congruence Θ of K extends to L, iff Θ is the reﬂection of conL (Θ). Figure 3.5 illustrates this in part. If a congruence Θ extends, then the congruence classes of Θ in K extend to congruence classes in L, but there may be congruence classes in L that are not such extensions.

3.3. Congruence-preserving extensions and variants

41

L K

Figure 3.5: A congruence extends. The extension map: ext : Con K → Con L maps a congruence Θ of K to the congruence conL (Θ) of L. The map ext is a {∨, 0}-homomorphism of Con K into Con L. In addition, ext preserves the zero, that is, ext is {0}-separating. To summarize: Lemma 3.12. Let K ≤ L be lattices. Then ext : Con K → Con L is a {0}separating join-homomorphism. The extension L of K is a congruence-reflecting extension (and the sublattice K of L is a congruence-reflecting sublattice) if every congruence of K extends to L. Utilizing the results of Sections 3.1 and 3.2, we can ﬁnd many equivalent formulations of the congruence-reflecting property for ﬁnite lattices: Lemma 3.13. Let L be a ﬁnite lattice, and K ≤ L. Then the following conditions are equivalent: (i) K is a congruence-reflecting sublattice of L. (ii) L is a congruence-reflecting extension of K. (iii) Let p and q be prime intervals in K; if p ⇒ q in L, then p ⇒ q in K. As an example, the reader may want to verify that any sublattice of a distributive lattice is congruence-reflecting.

42

3. Congruences

A much stronger concept—central to this book—is the following. Let K be a lattice. A lattice L is a congruence-preserving extension of K (or K is a congruence-preserving sublattice of L) if L is an extension and every congruence Θ of K has exactly one extension Θ to L satisfying ΘK = Θ. Of course, Θ = conL (Θ). It follows that Θ → conL (Θ) is an isomorphism between Con K and Con L.

K

L

K

L

Figure 3.6: Examples of congruence-preserving extensions. Two congruence-preserving extensions are shown in Figure 3.6, while Figure 3.7 shows two other extensions that are not congruence-preserving.

K

L

K

L

Figure 3.7: Examples of not congruence-preserving extensions. We can also obtain congruence-preserving extensions using gluing, based on the following result: Lemma 3.14. Let K and L be lattices, let F be a ﬁlter of K, and let I be an ideal of L. Let ψ be an isomorphism between F and I. Let G be the gluing of K and L over F and I with respect to ψ. If L is a congruence-preserving extension of I, then G is a congruence-preserving extension of K.

3.3. Congruence-preserving extensions and variants

43

If I and L are simple, then L is a congruence-preserving extension of I. So we obtain: Corollary 3.15. Let K, L, F , I, and ψ be given as above. If I and L are simple lattices, then G is a congruence-preserving extension of K. Lemma 3.16. Let the lattice L be an extension of the lattice K. Then L is a congruence-preserving extension of K iff the following two conditions hold: (i) re(ext Θ) = Θ, for any congruence Θ of K. (ii) ext(re Θ) = Θ, for any congruence Θ of L. We can say a lot more for ﬁnite lattices: Lemma 3.17. Let L be a ﬁnite lattice, and K ≤ L. Then L is a congruencepreserving extension of K iff the following two conditions hold: (a) Let p and q be prime intervals in K; if p ⇒ q in L, then p ⇒ q in K. (b) Let p be a prime interval of L. Then there exist a prime interval q in K such that p ⇔ q in L. Lemma 3.16.(ii) is very interesting by itself. It says that every congruence Θ of L is determined by its restriction to K. In other words, Θ = conL (ΘK). We will call such a sublattice K a congruence-determining sublattice. We can easily modify Lemma 3.17 to characterize congruence-determining sublattices in ﬁnite lattices: Lemma 3.18. Let L be a ﬁnite lattice L, and K ≤ L. Then K is a congruence-determining sublattice of L iff for any prime interval p in L, there is a prime interval q in K satisfying p ⇔ q in L. Of course, a congruence-preserving sublattice is always congruence-determining. In fact, a sublattice is congruence-preserving iff it is congruencereﬂecting and congruence-determining.

4 Chopped Lattices

The ﬁrst basic technique is the use of a chopped lattice, a ﬁnite meet-semilattice M, ∧ regarded as a partial algebra M, ∧, ∨, where ∨ is a partial operation. It turns out that the ideals of a chopped lattice form a lattice with the same congruence lattice as that of the chopped lattice. So to construct a ﬁnite lattice with a given congruence lattice it is enough to construct such a chopped lattice. The problem is how to ensure that the ideal lattice of the chopped lattice has some given properties. As an example, we will look at sectionally complemented lattices. Chopped lattices were introduced by G. Gr¨ atzer and H. Lakser, ﬁrst published in [24]. They were named in Gr¨ atzer and Schmidt [66] and generalized to the inﬁnite case in Gr¨ atzer and Schmidt [63].

4.1.

Basic deﬁnitions

An (n-ary) partial operation on a nonempty set A is a map from a subset of An to A. For n = 2, we call the partial operation binary. A partial algebra is a nonempty set A with partial operations deﬁned on A. A ﬁnite meet-semilattice M, ∧ may be regarded as a partial algebra, M, ∧, ∨, called a chopped lattice, where ∧ is an operation and ∨ is a partial operation: the least upper bound of a and b, provided that it exists. We can obtain an example of a chopped lattice by taking a ﬁnite lattice with unit, 1, and deﬁning M = L − {1}. The converse also holds: by adding a new unit 1 to a chopped lattice M , we obtain a ﬁnite lattice L, and chopping oﬀ the unit element, we get M back.

48

4. Chopped Lattices

A more useful example is obtained with merging. Let C and D be lattices such that J = C ∩ D is an ideal in both C and D. Then, with the natural ordering, Merge(C, D) = C ∪ D, called the merging of C and D, is a chopped lattice. Note that if a ∨ b = c in Merge(C, D), then either a, b, c ∈ C and a ∨ b = c in C or a, b, c ∈ D and a ∨ b = c in D. Among ﬁnite orders, chopped lattices are easy to spot. Lemma 4.1. Let M be a ﬁnite order and let Max be the set of maximal elements of M . If ↓ m is a lattice, for each m ∈ Max, and if m ∧ n exists, for all m, n ∈ Max, then M is a meet-semilattice. Proof. Indeed, for x, y ∈ M , x ∧ y = (x ∧m a) ∧a (y ∧n a), where m, n ∈ Max, x ≤ m, y ≤ n, a = m ∧ n, and ∧m , ∧n , ∧a denotes the meet in the lattice ↓ m, ↓ n, and ↓ a, respectively. It is easy to see that x ∧ y is the greatest lower bound of x and y in M . We deﬁne an equivalence relation Θ to be a congruence of a chopped lattice M as we deﬁned it for lattices in Section 1.3.3: we require that (SP∧ ) and (SP∨ ) hold, the latter with the proviso: whenever a ∨ c and b ∨ d exist. The set Con M of all congruence relations of M ordered by set inclusion is a lattice. Lemma 4.2. Let M be a chopped lattice and let Θ be an equivalence relation on M satisfying the following two conditions, for x, y, z ∈ M : (1) If x ≡ y (Θ); then x ∧ z ≡ y ∧ z (Θ). (2) If x ≡ y (Θ) and x ∨ z and y ∨ z exist, then x ∨ z ≡ y ∨ z (Θ). Then Θ is a congruence relation on M . Proof. Condition (1) states that Θ preserves ∧. Now let x, y, u, v ∈ S with x ≡ y (Θ) and u ≡ v (Θ); let x ∨ u and y ∨ v exist. Then x ≡ x ∧ y ≡ y (Θ) and (x ∧ y) ∨ u and (x ∧ y) ∨ v exist. Thus, by condition (2), x ∨ u ≡ (x ∧ y) ∨ u ≡ (x ∧ y) ∨ v ≡ y ∨ v

(Θ).

A nonempty subset I of the chopped lattice M is an ideal iff it is a down-set with the property: (Id) a, b ∈ I implies that a ∨ b ∈ I, provided that a ∨ b exists in M . The set Id M of all ideals of M ordered by set inclusion is a lattice. For I, J ∈ Id M , the meet is I ∩J, but the join is a bit more complicated to describe.

4.2. Compatible vectors of elements

49

Lemma 4.3. Let I and J be ideals of the chopped lattice M . Deﬁne U (I, J)0 = I ∪ J, U (I, J)i = { x | x ≤ u ∨ v, u, v ∈ U (I, J)i−1 }, for 0 < i < ω. Then I ∨J =

( U (I, J)i | i < ω ).

Proof. Deﬁne U = ( U (I, J)i | i < ω ). If K is an ideal of M , then I ⊆ K and J ⊆ K imply—by induction—that U ⊆ K. So it is suﬃcient to prove that U is an ideal of M . Obviously, U is a down-set. Also, U has property (Id), since if a, b ∈ U and a ∨ b exists in M , then a, b ∈ U (I, J)n , for some 0 < n < ω, and so a ∨ b ∈ U (I, J)n+1 ⊆ U . Most lattice concepts and notations for them will be used for chopped lattices without further explanation. In the literature, inﬁnite chopped lattices are also deﬁned, but we will not need them in this book.

4.2.

Compatible vectors of elements

Let M be a chopped lattice, and let Max(M ) (Max

if M is understood) be the set of maximal elements of M . Then M = ( id(m) | m ∈ Max ) and each id(m) is a (ﬁnite) lattice. A vector (associated with M ) is of the form im | m ∈ Max, where im ≤ m, for all m ∈ M . We order the vectors componentwise. With every ideal I of M , we can associate the vector im | m ∈ Max deﬁned by I ∩ id(m) = id(im ). Clearly, I = ( id(im ) | m ∈ M ). Such vectors are easy to characterize. Let us call the vector jm | m ∈ Max compatible if jm ∧ n = jn ∧ m, for all m, n ∈ Max. Lemma 4.4. Let M be a chopped lattice. (i) There is a one-to-one correspondence between ideals and compatible vectors of M . (ii) Given any vector g = gm | m ∈ Max, there is a smallest compatible vector g = im | m ∈ Max containing g. (iii) Let I and J be ideals of M , with corresponding compatible vectors im | m ∈ Max and jm | m ∈ Max. Then (a) I ≤ J in Id M iff im ≤ jm , for all m ∈ Max. (b) The compatible vector corresponding to I ∧J is im ∧jm | m ∈ Max.

50

4. Chopped Lattices

(c) Let a = im ∨ jm | m ∈ Max. Then the compatible vector corresponding to I ∨ J is a. Proof. (i) Let I be an ideal of M . Then im | m ∈ Max is compatible since im ∧ m ∧ n and in ∧ m ∧ n both generate the principal ideal I ∩ id(m) ∩ id(n), for all m, n ∈ Max. Conversely, let jm | m ∈ Max be compatible, and deﬁne I = ( id(jm ) | m ∈ Max ). Observe that (1)

I ∩ id(m) = id(jm ),

for m ∈ Max. Indeed, if x ∈ I ∩ id(m) and x ∈ id(jn ), for n ∈ Max, then x ≤ m ∧ jn = n ∧ jm (since jm | m ∈ Max is compatible), so x ≤ jm , that is, x ∈ id(jm ). The reverse inclusion is obvious. I is obviously a down-set. To verify property (Id) for I, let a, b ∈ I, and let us assume that a ∨ b exists in M . Then a ∨ b ≤ m, for some m ∈ Max, so a ≤ m and b ≤ m. By (1), we get a ≤ jm and b ≤ jm , so a ∨ b ≤ jm ∈ I. Since I is a down-set, it follows that a ∨ b ∈ I, verifying property (Id). (ii) Obviously, the vector m | m ∈ Max contains all other vectors and it is compatible. Since the componentwise meet of compatible vectors is compatible, the statement follows. (iii) is obvious since, by (ii), we are dealing with a closure system (see Section 1.2.2).

4.3.

Compatible vectors of congruences

Let M be a chopped lattice. With any congruence Θ of M , we can associate the reﬂection vector Θm | m ∈ Max, where Θm is the reﬂection (restriction) of Θ to id(m). The reﬂection Θm is a congruence of the lattice id(m). Let Φm be a congruence of id(m), for all m ∈ Max. The congruence vector Φm | m ∈ Max is called compatible if Φm restricted to id(m ∧ n) is the same as Φn restricted to id(m ∧ n), for m, n ∈ Max. Obviously, a reﬂection vector is compatible. The converse also holds. Lemma 4.5. Let Φm | m ∈ Max be a compatible congruence vector of a chopped lattice M . Then there is a unique congruence Θ of M such that the reﬂection vector of Θ agrees with Φm | m ∈ Max. Proof. Let Φm | m ∈ Max be a compatible congruence vector. We deﬁne a binary relation Θ on M as follows:

4.3. Compatible vectors of congruences

51

Let m, n ∈ Max. For x ∈ id(m) and y ∈ id(n), let x ≡ y (Θ) iff x ≡ x ∧ y (Φm ) and y ≡ x ∧ y (Φn ). Obviously, Θ is reﬂexive and symmetric. To prove transitivity, let m, n, k ∈ Max, and let x ∈ id(m), y ∈ id(n), z ∈ id(k); let x ≡ y (Θ) and y ≡ z (Θ), that is, (2)

x≡x∧y

(Φm ),

(3)

y ≡x∧y

(Φn ),

(4)

y ≡y∧z

(Φn ),

(5)

z ≡y∧z

(Φk ).

Then meeting the congruence (4) with x (in the lattice id(n)), we get x∧y ≡x∧y∧z

(6)

(Φn ),

and from (3), by meeting with z, we obtain y∧z ≡x∧y∧z

(7)

(Φn ).

Since x ∧ y and x ∧ y ∧ z ∈ id(m), by compatibility, (6) implies that x∧y ≡x∧y∧z

(8)

(Φm ).

Now (2) and (8) yield (9)

x≡x∧y∧z

(Φm ).

z ≡x∧y∧z

(Φk ).

Similarly, (10)

Φm is a lattice congruence on id(m) and x ∧ y ∧ z ≤ x ∧ z ≤ x, so (11)

x≡x∧z

(Φm ).

z ≡x∧z

(Φk ).

Similarly, (12)

Equation (11) and (12) yield that x ≡ z (Θ), proving transitivity. (SP∧ ) is easy: let x ∈ id(m), y ∈ id(n), z ∈ M ; if x ≡ y (Θ), then x ∧ z ≡ y ∧ z (Θ) because x ∧ z ≡ x ∧ y ∧ z (Φm ) and y ∧ z ≡ x ∧ y ∧ z (Φn ). Finally, we verify (SP∨ ). Let x ≡ y (Θ) and z ∈ M , and let us assume that x ∨ z and y ∨ z exist. Then there are p, q ∈ Max such that x ∨ z ∈ id(p) and y ∨ z ∈ id(q). By compatibility, x ≡ x ∧ y (Φp ), so x ∨ z ≡ (x ∧ y) ∨ z (Φp ). Since (x ∧ y) ∨ z ≤ (x ∨ z) ∧ (y ∨ z) ≤ x ∨ z, we also have x ∨ z ≡ (x ∨ z) ∧ (y ∨ z)

(Φp ).

y ∨ z ≡ (x ∨ z) ∧ (y ∨ z)

(Φq ).

Similarly, The last two displayed equations show that x ∨ z ≡ y ∨ z (Θ).

52

4.4.

4. Chopped Lattices

From the chopped lattice to the ideal lattice

The map m → id(m) embeds the chopped lattice M with zero into the lattice Id M , so we can regard Id M as an extension. It is, in fact, a congruence-preserving extension (Gr¨ atzer and Lakser [34], proof ﬁrst published in [23]): Theorem 4.6. Let M be a chopped lattice. Then Id M is a congruence-preserving extension of M . Proof. Let Θ be a congruence relation of M . If I, J ∈ Id M , deﬁne I ≡ J (Θ)

iff I/Θ = J/Θ.

Obviously, Θ is an equivalence relation. Let I ≡ J (Θ), N ∈ Id M , and x ∈ I ∩N . Then x ≡ y (Θ), for some y ∈ J, and so x ≡ x∧y (Θ) and x∧y ∈ J ∩N . This shows that (I ∩ N )/Θ ⊆ (J ∩ N )/Θ. Similarly, (J ∩ N )/Θ ⊆ (I ∩ N )/Θ, so I ∩ N ≡ J ∩ N (Θ). To prove I ∨ N ≡ J ∨ N (Θ), by symmetry, it is suﬃcient to verify that I ∨ N ⊆ (J ∨ N )/Θ. By Lemma 4.3, this is equivalent to proving that Un ⊆ (J ∨ N )/Θ, for n < ω. This is obvious for n = 0. Now assume that Un−1 ⊆ (J ∨ N )/Θ and let x ∈ Un . Then x ≤ t1 ∨ t2 , for some t1 , t2 ∈ Un−1 . Thus t1 ≡ u1 (Θ) and t2 ≡ u2 (Θ), for some u1 , u2 ∈ J ∨ N , and so t1 ≡ t1 ∧ u1 (Θ) and t2 ≡ t2 ∧ u2 (Θ). Observe that t1 ∨ t2 is an upper bound for {t2 ∧ u1 , t2 ∧ u2 }; consequently, (t1 ∧ u1 ) ∨ (t2 ∧ u2 ) exists. Therefore, t1 ∨ t2 ≡ (t1 ∧ u1 ) ∨ (t2 ∧ u2 )

(Θ).

Finally, x ≡ x ∧ (t1 ∨ t2 ) = x ∧ ((t1 ∧ u1 ) ∨ (t2 ∧ u2 ))

(Θ),

and x ∧ ((t1 ∧ u1 ) ∨ (t2 ∧ u2 )) ∈ J ∨ N. Thus x ∈ (J ∨ N )/Θ, completing the induction, verifying that Θ is a congruence relation of Id M . If a ≡ b (Θ) and x ∈ id(a), then x ≡ x ∧ b (Θ). Thus id(a) ⊆ id(b)/Θ. Similarly, id(b) ⊆ id(a)/Θ, and so id(a) ≡ id(b) (Θ). Conversely, if id(a) ≡ id(b) (Θ), then a ≡ b1 (Θ) and a1 ≡ b (Θ), for some a1 ≤ a and b1 ≤ b. Forming the join of these two congruences, we get a ≡ b (Θ). Thus Θ has all the properties required by Lemma 4.3. To show the uniqueness, let Φ be a congruence relation of Id M satisfying id(a) ≡ id(b) (Φ) iff a ≡ b (Θ). Let I, J ∈ Id M , I ≡ J (Φ), and x ∈ I. Then id(x) ∩ I ≡ id(x) ∩ J id(x) ∩ I = id(x), id(x) ∩ J = id(y),

(Φ),

4.5. Sectional complementation

53

for some y ∈ J. Thus id(x) ≡ id(y) (Φ), and so x ≡ y (Θ), proving that I ⊆ J/Θ. Similarly, J ⊆ I/Θ, and so I ≡ J (Θ). Conversely, if I ≡ J (Θ), then take all congruences of the form x ≡ y (Θ), x ∈ I, y ∈ J. By our assumption regarding Φ, we get the congruence id(x) ≡ id(y) (Φ), and by our deﬁnition of Θ, the join of all these congruences yields I ≡ J (Θ). Thus Φ = Θ. This result is very useful. It means that in order to construct a ﬁnite lattice L to represent a given ﬁnite distributive lattice D as a congruence lattice, it is suﬃcient to construct a chopped lattice M with Con M ∼ = D, since Con M ∼ = Con(Id M ) = Con L, where L = Id M , and L is a ﬁnite lattice. This result also allows us to construct congruence-preserving extensions. Corollary 4.7. Let M = Merge(A, B) be a chopped lattice with A = id(a) and B = id(b). If a ∧ b > 0 and B is simple, then Id M is a congruence-preserving extension of A. Proof. Let Θ be a congruence of A. Then Θ, Ψ is a compatible congruence vector iff ω, if Θ is discrete on [0, a ∧ b]; Ψ= ι, otherwise. So Ψ is determined by Θ and the statement follows.

4.5.

Sectional complementation

We introduce sectionally complemented chopped lattices as we did for lattices in Section 2.1. We illustrate the use of compatible vectors with two results on sectionally complemented chopped lattices. The ﬁrst result is from Gr¨ atzer and Schmidt [69]. Lemma 4.8 (Atom Lemma). Let M be a chopped lattice with two maximal elements m1 and m2 . We assume that id(m1 ) and id(m2 ) are sectionally complemented lattices. If p = m1 ∧ m2 is an atom, then Id M is sectionally complemented. Proof. Figure 4.1 illustrates the setup. To show that Id M is sectionally complemented, let I ⊆ J be two ideals of M , represented by the compatible vectors i1 , i2 and j1 , j2 , respectively. Let s1 be the sectional complement of i1 in j1 and let s2 be the sectional complement of i2 in j2 . If p ∧ s1 = p ∧ s2 , then s1 , s2 is a compatible vector, representing an ideal S that is a sectional complement of I in J. Otherwise, without loss of generality, we can assume that p ∧ s1 = 0 and p ∧ s2 = p. Since

54

4. Chopped Lattices

m2 m1 p

M 0 Figure 4.1: Atom Lemma illustrated. id(m2 ) is sectionally complemented, there is a sectional complement s2 of p in [0, s2 ]. Then s1 , s2 satisﬁes p ∧ s1 = p ∧ s2 (= 0), and so it is compatible; therefore, s1 , s2 represents an ideal S of M . Obviously, I ∧ S = {0}. From p ∧ s2 = p, it follows that p ≤ s2 ≤ j2 . Since J is an ideal and j2 ∧ p = p, it follows that j1 ∧ p = p, that is, p ≤ j1 . Obviously, I ∨ S ⊆ J. So to show that I ∨ S = J, it is suﬃcient to verify that j1 , j2 ∈ I ∨ S. Evidently, j1 = i1 ∨ s1 ∈ I ∨ S. Note that p ≤ j1 = i1 ∨ s1 ∈ I ∨ S. Thus p, s2 , i2 ∈ I ∨ S, and therefore p ∨ s2 ∨ i2 = (p ∨ s2 ) ∨ i2 = s2 ∨ i2 = j2 ∈ I ∨ S. The second result (Gr¨ atzer, Lakser, and Roddy [44]) shows that the ideal lattice of a sectionally complemented chopped lattice is not always sectionally complemented. Theorem 4.9. There is a sectionally complemented chopped lattice M whose ideal lattice Id M is not sectionally complemented. Proof. Let M be the chopped lattice of Figure 4.2, where L1 = id(m1 ) and L2 = id(m2 ). Note that p is meet-irreducible in id(m2 ) and q is meetirreducible in id(m1 ). The unit element of the ideal lattice of M is the compatible vector m1 , m2 . We show that the compatible vector a, b has no complement in the ideal lattice of M . Assume, to the contrary, that the compatible vector s, t is a complement of a, b. Since a, b ≤ a ∨ u, m2 , a compatible vector, s, t a ∨ u, m2 , that is, (13)

s a ∨ u.

Similarly, by considering m1 , b ∨ u, we conclude that (14)

t b ∨ u.

4.5. Sectional complementation

55

m2

m1

q

p c

d u b

a p

q

L1

L2 0

Figure 4.2: The chopped lattice M . Now a, b ≤ p , q , which is a compatible vector. Thus s, t p , q , and so either s p or t q . Without loss of generality, we may assume that s p . It then follows by (13) that s can be only c or m1 . Then, since s ∧ a = 0, we conclude that s = c. Thus s ∧ u = p, and so t ∧ u = p. But p is meet-irreducible in L2 . Thus t = p ≤ b ∨ u, contradicting (14). This result illustrates that the Atom Lemma (Lemma 4.8) cannot be extended to the case where [0, m1 ∧ m2 ] is a four-element boolean lattice.

5 Boolean Triples

In Part IV we construct congruence-preserving extensions of ﬁnite lattices, extensions with special properties, such as sectionally complemented, semimodular, and so on. It is easy to construct a proper congruence-preserving extension of a ﬁnite lattice. In the early 1990s, G. Gr¨ atzer and E. T. Schmidt raised the question in [66] whether every lattice has a proper congruence-preserving extension. (See also Gr¨ atzer and Schmidt [63].) It took almost a decade for the answer to appear in Gr¨ atzer and Wehrung [81]. For inﬁnite lattices, the aﬃrmative answer was provided by their construction: boolean triples, which is described in this chapter. It is interesting that boolean triples also provide a very important tool for ﬁnite lattices.

5.1.

The general construction

In this section I describe a congruence-preserving extension of a (ﬁnite) lattice L, introduced in Gr¨ atzer and Wehrung [81]. We will see that this generalizes a construction of Schmidt [106] for bounded distributive lattices. For a lattice L, let us call the triple x, y, z ∈ L3 boolean iff (F stands for “Fixed point deﬁnition”)

(F)

x = (x ∨ y) ∧ (x ∨ z), y = (y ∨ x) ∧ (y ∨ z), z = (z ∨ x) ∧ (z ∨ y). Note that by Lemma 2.6 if (F) holds then sub({x, y, z}) is boolean.

58

5. Boolean Triples

(F) is a “Fixed point deﬁnition” because the triple x, y, z satisﬁes (F) iff p(x, y, z) = x, y, z, where p = (x ∨ y) ∧ (x ∨ z), (y ∨ x) ∧ (y ∨ z), (z ∨ x) ∧ (z ∨ y). We denote by M3 [L] ⊆ L3 the order of boolean triples of L (ordered as a suborder of L3 , that is, componentwise). If we apply the construction to an interval [a, b] of L, we write M3 [a, b] for M3 [[a, b]]. Observe that any boolean triple x, y, z ∈ L3 satisﬁes (B stands for “balanced”) x ∧ y = y ∧ z = z ∧ x.

(B)

Indeed, if x, y, z is boolean, then x ∧ y = y ∧ z = z ∧ x = (x ∨ y) ∧ (y ∨ z) ∧ (z ∨ x). We call such triples balanced. Here are some of the basic properties of boolean triples: Lemma 5.1. Let L be a lattice. (i) x, y, z ∈ L3 is boolean iﬀ there is a triple u, v, w ∈ L3 satisfying (E stands for “existential”) x = u ∧ v, (E)

y = u ∧ w, z = v ∧ w.

(ii) M3 [L] is a closure system in L3 . For x, y, z ∈ L3 , the closure is x, y, z = (x ∨ y) ∧ (x ∨ z), (y ∨ x) ∧ (y ∨ z), (z ∨ x) ∧ (z ∨ y). (iii) If L has 0, then the suborder { x, 0, 0 | x ∈ L } is a sublattice of M3 [L] and ϕ : x → x, 0, 0 is an isomorphism between L and this sublattice. (iv) If L is bounded, then M3 [L] has a spanning M3 , that is, a {0, 1}-sublattice isomorphic to M3 , namely, {0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1}. Remark. (E) is the “Existential deﬁnition” because x, y, z is boolean iﬀ there exists a triple u, v, w satisfying (E).

5.1. The general construction

59

Proof. (i) If x, y, z is boolean, then u = x ∨ y, v = x ∨ z, and w = y ∨ z satisfy (E). Conversely, if there is a triple u, v, w ∈ L3 satisfying (E), then by Lemma 2.6, the sublattice generated by x, y, and z is isomorphic to a quotient of B3 and x, y, and z are the images of the three atoms of B3 . Thus (x ∨ y) ∧ (x ∨ z) = x, the ﬁrst equation in (F). The other two equations are proved similarly. (ii) M3 [L] = ∅; for instance, for all x ∈ L, the diagonal element x, x, x ∈ M3 [L]. For x, y, z ∈ L3 , deﬁne u = x ∨ y, v = x ∨ z, and w = y ∨ z. Set x1 = u ∧ v, y1 = u ∧ w, and z1 = v ∧ w. Then x1 , y1 , z1 is boolean by (i) and x, y, z ≤ x1 , y1 , z1 in L3 . Now if x, y, z ≤ x2 , y2 , z2 in L3 and x2 , y2 , z2 is boolean, then x2 = (x2 ∨ y2 ) ∧ (x2 ∨ z2 ) ≥ (x ∨ y) ∧ (x ∨ z)

(by (F)) (by x, y, z ≤ x2 , y2 , z2 )

= u ∧ v = x1 , and similarly, y2 ≥ y1 , z2 ≥ z1 . Thus x2 , y2 , z2 ≥ x1 , y1 , z1 , and so x1 , y1 , z1 is the smallest boolean triple containing x, y, z. (iii) and (iv) are obvious. M3 [L] is diﬃcult to draw in general. Figure 5.1 shows the diagram of M3 [C3 ] with the three-element chain C3 = {0, a, 1}. A three-dimensional rendering of M3 [C3 ] is posted on my Web site. If C is an arbitrary bounded chain, with bounds 0 and 1, it is easy to picture M3 [C], as sketched in Figure 5.1. The element x, y, z ∈ C 3 is boolean iﬀ it is of the form x, y, y, or y, x, y, or y, y, x, where y ≤ x in C. So the diagram is made up of three isomorphic “ﬂaps” overlapping on the diagonal

1, 1, 1 1, 1, a 1, 0, 0

a, a, a

a, 0, 0

a, 1, 1

1, a, 1

0, a, 0

0, 0, 1

0, 1, 0

0, 0, a 0, 0, 0

Figure 5.1: The lattice M3 [C3 ] with a sketch.

60

5. Boolean Triples

(the elements of the form x, x, x). Two of the ﬂaps form the “base”, a planar lattice: C 2 , the third one (shaded) comes up out of the plane pointing in the direction of the viewer. We get some more examples of M3 [L] from the following observation: Lemma 5.2. Let the lattice L have a direct product decomposition: L = L1 × L2 . Then M3 [L] ∼ = M3 [L1 ] × M3 [L2 ]. Proof. This is obvious, since x1 , y1 , x2 , y2 , x3 , y3 is a boolean triple iff x1 , x2 , x3 and y1 , y2 , y3 both are (xi ∈ L1 and yi ∈ L2 , for i = 1, 2, 3).

5.2.

The congruence-preserving extension property

Let L be a nontrivial lattice with zero and let ϕ : x → x, 0, 0 ∈ M3 [L] be an embedding of L into M3 [L]. Here is the main result of Gr¨ atzer and Wehrung [81]: Theorem 5.3. M3 [L] is a congruence-preserving extension of Lϕ. The next two lemmas prove this theorem. For a congruence Θ of L, let Θ3 denote the congruence of L3 deﬁned componentwise. Let M3 [Θ] be the reﬂection of Θ3 to M3 [L]. Lemma 5.4. M3 [Θ] is a congruence relation of M3 [L]. Proof. M3 [Θ] is obviously an equivalence relation on M3 [L]. Since M3 [L] is a meet-subsemilattice of L3 , it is clear that M3 [Θ] satisﬁes (SP∧ ). To verify (SP∨ ) for M3 [Θ], let x1 , y1 , z1 , x2 , y2 , z2 ∈ M3 [L], let x1 , y1 , z1 ≡ x2 , y2 , z2 (M3 [Θ]), and let u, v, w ∈ M3 [L]. Set xi , yi , zi = xi , yi , zi ∨ u, v, w (the join formed in M3 [L]), for i = 1, 2. Then, using Lemma 5.1.(ii) for x1 ∨ u, y1 ∨ v, and z1 ∨ w, we obtain that x1 = (x1 ∨ u ∨ y1 ∨ v) ∧ (x1 ∨ u ∨ z1 ∨ w)

≡ (x2 ∨ u ∨ y2 ∨ v) ∧ (x2 ∨ u ∨ z2 ∨ w) = x2

(M3 [Θ]),

and similarly, y1 ≡ y2 (M3 [Θ]), z1 ≡ z2 (M3 [Θ]), hence x1 , y1 , z1 ∨ u, v, w ≡ x2 , y2 , z2 ∨ u, v, w (M3 [Θ]).

5.2. The congruence-preserving extension property

61

It is obvious that M3 [Θ] restricted to Lϕ is Θϕ. Lemma 5.5. Every congruence of M3 [L] is of the form M3 [Θ], for a suitable congruence Θ of L. Proof. Let Φ be a congruence of M3 [L], and let Θ denote the congruence of L obtained by restricting Φ to the sublattice Lϕ = { x, 0, 0 | x ∈ L } of M3 [L], that is, for x, y ∈ L, x ≡ y (Θ) iff x, 0, 0 ≡ y, 0, 0 (Φ). We prove that Φ = M3 [Θ]. To show that Φ ⊆ M3 [Θ], let x1 , y1 , z1 ≡ x2 , y2 , z2 (Φ).

(1)

Meeting the congruence (1) with 1, 0, 0 yields x1 , 0, 0 ≡ x2 , 0, 0 (Φ), and so x1 , 0, 0 ≡ x2 , 0, 0 (M3 [Θ]).

(2)

Meeting the congruence (1) with 0, 1, 0 yields 0, y1 , 0 ≡ 0, y2 , 0 (Φ).

(3) Since

0, y1 , 0 ∨ 0, 0, 1 = 0, y1 , 1 = y1 , y1 , 1, and y1 , y1 , 1 ∧ 1, 0, 0 = y1 , 0, 0, and similarly for 0, y2 , 0, joining the congruence (3) with 0, 1, 0 and then meeting with 1, 0, 0, yields y1 , 0, 0 ≡ y2 , 0, 0 (Φ), and so 0, y1 , 0 ≡ 0, y2 , 0 (M3 [Θ]).

(4) Similarly, (5)

0, 0, z1 ≡ 0, 0, z2 (M3 [Θ]).

Joining the congruences (2), (4), and (5), we obtain (6)

x1 , y1 , z1 ≡ x2 , y2 , z2 (M3 [Θ]),

proving that Φ ⊆ M3 [Θ].

62

5. Boolean Triples

To prove the converse, M3 [Θ] ⊆ Φ, take x1 , y1 , z1 ≡ x2 , y2 , z2 (M3 [Θ])

(7)

in M3 [L]; equivalently, (8) (9) (10)

x1 , 0, 0 ≡ x2 , 0, 0 (Φ), y1 , 0, 0 ≡ y2 , 0, 0 (Φ), z1 , 0, 0 ≡ z2 , 0, 0 (Φ)

in M3 [L]. Joining the congruence (9) with 0, 0, 1 and then meeting the result with 0, 1, 0, we get (as in the computation following (3)): (11)

0, y1 , 0 ≡ 0, y2 , 0 (Φ).

Similarly, from (10), we conclude that (12)

0, 0, z1 ≡ 0, 0, z2 (Φ)

Finally, joining the congruences (8), (11), and (12), we get (13)

x1 , y1 , z1 ≡ x2 , y2 , z2 (Φ),

that is, M3 [Θ] ⊆ Φ. This completes the proof of this lemma.

5.3.

The distributive case

Let D be a bounded distributive lattice. In 1974 (25 years before the publication of Gr¨ atzer and Wehrung [81]), Schmidt [106] deﬁned M3 [D] as the set of balanced triples x, y, z ∈ D3 (deﬁned in Section 5.1), regarded as a suborder of D3 . Schmidt proved the following result: Theorem 5.6. Let D be a bounded distributive lattice. Then M3 [D] (the set of balanced triples x, y, z ∈ D3 ) is a modular lattice. The map ϕ : x → x, 0, 0 ∈ M3 [D] is an embedding of D into M3 [D], and M3 [D] is a congruence-preserving extension of Dϕ. Proof. Examining Figure 2.7 (page 27), we immediately see that in a distributive lattice D, conditions (B) and (F) are equivalent, so M3 [D] is the boolean triple construction. Therefore, this result follows from the results in Sections 5.1 and 5.2, except for the modularity. A direct computation of this is not so easy—although entertaining. However, we can do it without computation. Observe that it is enough to prove modularity for a ﬁnite D. Now if D is ﬁnite, then by Lemma 5.2 and Corollary 2.18, M3 [D] can be embedded into M3 [Bn ] ∼ = (M3 [C2 ])n ∼ = M3n , a modular lattice; hence M3 [D] is modular.

5.4. Two interesting intervals

5.4.

63

Two interesting intervals

Let L be a bounded lattice. For an arbitrary a ∈ L, consider the following interval M3 [L, a] of M3 [L] (illustrated in Figure 5.2): M3 [L, a] = [0, a, 0, 1, 1, 1] ⊆ M3 [L].

1, 1, 1

1, a, a 1, 0, 0

M3 [L, a] 0, 1, 0

B

K

M3 [L]

0, a, 0

0, 0, 1

0, 0, 0 Figure 5.2: The shaded area: The lattice M3 [L, a]. Deﬁne the map (with image B = Lϕa in Figure 5.2): ϕa : x → x, a, x ∧ a. Lemma 5.7. ϕa is an embedding of L into M3 [L, a]. Proof. ϕa is obviously one-to-one and meet-preserving. It also preserves the join because x ∨ y, a, (x ∧ a) ∨ (y ∧ a) = x ∨ y, a, (x ∨ y) ∧ a, for x, y ∈ L. The following result of Gr¨ atzer and Schmidt [71] is a generalization of Theorem 5.3: Theorem 5.8. M3 [L, a] is a congruence-preserving extension of Lϕa . The proof is presented in the following lemmas. Lemma 5.9. Let x, y, z ∈ M3 [L, a]. Then a ≤ y and (14)

x ∧ a = z ∧ a.

64

5. Boolean Triples

Proof. Since x, y, z is boolean, it is balanced, so x ∧ y = z ∧ y. Therefore, x ∧ a = (x ∧ y) ∧ a = (z ∧ y) ∧ a = z ∧ a, as claimed. We need an easy decomposition statement for the elements of M3 [L, a]. Let us use the notation (B and K are marked in Figure 5.2): B = { x, a, x ∧ a | x ∈ L } (= Lϕa ), K = { 0, x, 0 | x ∈ L, x ≥ a }, J = { x ∧ a, a, x | x ∈ L }. Lemma 5.10. Let v = x, y, z ∈ M3 [L, a]. Then v has a decomposition in M3 [L, a]: (15)

v = vB ∨ vK ∨ vJ ,

where (16)

vB = x, y, z ∧ 1, a, a = x, a, x ∧ a ∈ B,

(17)

vK = x, y, z ∧ 0, 1, 0 = 0, y, 0 ∈ K, vJ = x, y, z ∧ a, a, 1 = z ∧ a, a, z ∈ J.

(18)

Proof. (16) follows from (14). By symmetry, (18) follows, and (17) is trivial. Finally, the right side of (15) componentwise joins to the left side—in view of (14). We can now describe the congruences of M3 [L, a]. Lemma 5.11. Let Φ be a congruence of M3 [Θ, a] and let v, w ∈ M3 [L, a]. Then (19)

v≡w

(Φ),

iff (20)

vB ≡ wB

(Φ),

(21)

vK ≡ wK

(Φ),

(22)

vJ ≡ wJ

(Φ).

Proof. (19) implies (20) by (16). Similarly, for (21) and (22). Conversely, (20)–(22) imply (19) by (15). Lemma 5.12. For a congruence Θ of L, let M3 [Θ, a] be the reﬂection of Θ3 to M3 [L, a]. Then M3 [Θ, a] is a congruence of M3 [L, a], and every congruence of M3 [L, a] is of the form M3 [Θ, a], for a unique congruence Θ of L.

5.4. Two interesting intervals

65

Proof. It follows from Lemma 5.4 that M3 [Θ, a] is a congruence of M3 [L, a]. Let v = x, y, z, v = x , y , z ∈ M3 [L, a]. Let Φ be a congruence of M3 [L, a], and let Θ be the reﬂection of Φ to L with respect to the embedding ϕa . By Lemma 5.11, v ≡ v (Φ) iff (20)–(22) hold. Note that vB , vB ∈ Lϕa , so (20) is equivalent to vB ≡ vB (Θ). Now consider p(x) = (x ∨ 0, 1, 0) ∧ 1, a, a.

Then p(x ∧ a, a, x) = x, 1, x ∧ 1, a, a = x, a, x ∧ a. So (22) implies that p(vJ ) ≡ p(vJ ) (Φ), and symmetrically. Thus (22) is equivalent to

that is, to

p(vJ ) ≡ p(vJ )

(Φ),

p(vJ ) ≡ p(vJ )

(Θ),

since p(vJ ), p(vJ ) ∈ Lϕa . Now consider q(x) = (x ∨ a, a, 1) ∧ 1, a, a. Then q(0, x, 0) = x, a, x ∧ a, for x ≥ a. So q(vK ) ≡ q(vK ) (Φ), that is, q(vK ) ≡ q(vK ) (Θ). Finally, deﬁne r(x) = (x ∨ a, a, 1) ∧ 0, 1, 0. Then q(x, x ∧ a, a) = 0, x, 0. So q(vB ) ≡ q(vB ) (Φ). From these it follows ) (Φ) and q(vK ), q(vK )∈ that vK ≡ vK (Φ) is equivalent to q(vK ) ≡ q(vK Lϕa , so the latter is equivalent to q(vK ) ≡ q(vK ) (Θ). We conclude that the congruence v ≡ v (Φ) in M3 [L, a] is equivalent to the following three congruences in L: vB ≡ vB p(vJ ) ≡ p(vJ )

q(vK ) ≡

q(vK )

(Θ), (Θ), (Θ),

concluding the proof of the lemma. In Chapter 16, we need a smaller interval introduced in Gr¨ atzer and Schmidt [75]; namely, for a, b ∈ L with a < b, we introduce the interval M3 [L, a, b] of M3 [L]: M3 [L, a, b] = [0, a, 0, 1, b, b] ⊆ M3 [L]. Again, ϕa : x → x, a, x ∧ a

66

5. Boolean Triples

1, 1, 1 1, b, b 1, a, a 1, 0, 0

Da,b 0, 1, 0 0, 0, 1

0, b, 0

Ia,b

M3 [L]

0, a, 0

0, 0, 0 Figure 5.3: The shaded area: The lattice M3 [L, a, b]. is a (convex) embedding of L into M3 [L, a, b]. (Note that if L is bounded, then M3 [L, a] = M3 [L, a, 1].) Using the notation (illustrated in Figure 5.4; the black-ﬁlled elements form J): B = { x, a, x ∧ a | x ∈ L } (= Lϕa ), Ia,b = [0, a, 0, 0, b, 0], J = { x ∧ a, a, x | x ≤ b }. We can now generalize Lemmas 5.10–5.12: Lemma 5.13. Let v = x, y, z ∈ M3 [L, a, b]. Then v has a decomposition in M3 [L, a, b]: v = vB ∨ vIa,b ∨ vJ , where vB = x, y, z ∧ 1, a, a = x, a, x ∧ a ∈ B, vIa,b = x, y, z ∧ 0, b, 0 = 0, y, 0 ∈ Ia,b , vJ = x, y, z ∧ a, a, b = z ∧ a, a, z ∈ J. Lemma 5.14. Let Φ be a congruence of M3 [L, a, b] and let v, w ∈ M3 [L, a, b]. Then v ≡ w (Φ) iff (23) (24) (25)

vB ≡ wB

(Φ),

vIa,b ≡ wIa,b (Φ), vJ ≡ wJ (Φ).

5.4. Two interesting intervals

67

1, b, b

Fa,b

1, a, a

a, a,b

B

0, b, 0

J Ia,b 0, a, 0

Figure 5.4: The lattice M3 [L, a, b]. Lemma 5.15. For a congruence Θ of L, let M3 [Θ, a, b] be the restriction of Θ3 to M3 [L, a, b]. Then M3 [Θ, a, b] is a congruence of M3 [L, a, b], and every congruence of M3 [L, a, b] is of the form M3 [Θ, a, b], for a unique congruence Θ of L. It follows that ϕa is a congruence-preserving convex embedding of L into M3 [L, a, b]. We will use the notation Fa,b = [0, b, 0, 1, b, b] = { x, b, x ∧ b | x ∈ L }. The following two observations are trivial. Lemma 5.16. (i) Ia,b is an ideal of M3 [L, a, b] and Ia,b is isomorphic to the interval [a, b] of L. (ii) Fa,b is a ﬁlter of M3 [L, a, b] and Fa,b is isomorphic to L. We can say a lot more about Fa,b . But ﬁrst we need another lemma: Lemma 5.17. Let L be a lattice, let [u, v] and [u , v ] be intervals of L, and let ϕ : [u, v] → [u , v ] be an isomorphism between these two intervals. Let ϕ and ϕ−1 be algebraic in L. Then L is a congruence-preserving extension of [u, v] iff it is a congruence-preserving extension of [u , v ].

68

5. Boolean Triples

Proof. Let us assume that L is a congruence-preserving extension of [u , v ]. Let Θ be a congruence relation of [u, v] and let Θϕ be the image of Θ under ϕ. Since ϕ is an isomorphism, it follows that Θϕ is a congruence of [u , v ], and so Θϕ has a unique extension to a congruence Φ of L. We claim that Φ extends Θ to L and extends it uniquely. 1. Φ extends Θ. Let x ≡ y (Θ). Then xϕ ≡ yϕ (Θϕ), since ϕ is an isomorphism. By deﬁnition, Φ extends Θϕ, so xϕ ≡ yϕ (Φ). Since ϕ−1 is algebraic, the last congruence implies that x ≡ y (Φ). Conversely, let x ≡ y (Φ) and x, y ∈ [u, v]. Then xϕ, yϕ ∈ [u , v ]; since ϕ is algebraic, it follows that xϕ ≡ yϕ (Φ). Since Φ extends Θϕ, we conclude that xϕ ≡ yϕ (Θϕ). Using that ϕ−1 is an isomorphism, we obtain that x ≡ y (Θ), verifying the claim. 2. Φ extends Θ uniquely. Let Ψ extend Θ to L. As in the previous paragraph—mutatis mutandis— we conclude that Ψ extends Θϕ to L, hence Ψ = Φ, proving the uniqueness and the claim. By symmetry, the lemma is proved. We have already observed in Lemma 5.12 that M3 [L, a, b] is a congruencepreserving convex extension of [0, a, 0, 1, a, a] (= Lϕa ). Since the isomorphism x → x ∨ 0, b, 0 between [0, a, 0, 1, a, a] and [0, b, 0, 1, b, b] = Fa,b is algebraic, and so is the inverse x → x ∧ 1, a, a, from Lemma 5.17 we conclude the following: Corollary 5.18. The lattice M3 [L, a, b] is a congruence-preserving convex extension of the ﬁlter Fa,b . We summarize our results (Gr¨ atzer and Schmidt [75]): Lemma 5.19. Let L be a bounded lattice, and let a, b ∈ L with a < b. Then there exists a bounded lattice La,b (with bounds 0a,b and 1a,b ) and ua,b , va,b ∈ La,b , such that the following conditions are satisﬁed: (i) va,b is a complement of ua,b . (ii) Fa,b = [va,b , 1a,b ] ∼ = L. (iii) Ia,b = [0a,b , va,b ] ∼ = [a, b]. (iv) La,b is a congruence-preserving (convex ) extension of [0a,b , ua,b ] and of [va,b , 1a,b ].

5.4. Two interesting intervals

69

(v) The congruences on Ia,b and Fa,b are synchronized, that is, if Θ is a congruence on L, Θ is the extension of Θ to La,b (we map Θ to Fa,b under the isomorphism, and then by (iv) we uniquely extend it to La,b ), and x, y ∈ [a, b], then we can denote by xFa,b , yFa,b ∈ Fa,b the images of x, y in Fa,b and by xIa,b , yIa,b ∈ Ia,b the images of x, y in Ia,b ; synchronization means that xFa,b ≡ yFa,b (Θ) iff xIa,b ≡ yIa,b (Θ). Proof. Of course, La,b = M3 [L, a, b], 0a,b = 0, a, 0, 1a,b = 1, b, b, ua,b = 1, a, a, and va,b = 0, b, 0. The lattice La,b is illustrated with L = C5 in Figure 5.5, the ﬁve-element chain, a is the atom and b is the dual atom of C5 . This ﬁgure is the same as Figure 5.4, only the notation is changed.

1a,b

Fa,b ua,b

va,b

Ia,b

0a,b Figure 5.5: The lattice La,b .

6 Cubic Extensions

In this chapter, for a ﬁnite lattice K, we introduce an extension Cube K with the following properties: (i) The lattice K is a congruence-reflecting sublattice of Cube K. (ii) Con(Cube K) is boolean. (iii) The minimal extension of the meet-irreducible congruences are the dual atoms of Con(Cube K); their ordering is “ﬂattened.”

6.1.

The construction

Let K be a ﬁnite lattice. Following Gr¨ atzer and Schmidt [69], for every meet-irreducible congruence Ψ of K (in formula, Ψ ∈ ConM K), we form the quotient lattice K/Ψ, and extend it to a ﬁnite simple lattice Simp KΨ with zero 0Ψ and unit 1Ψ , using Lemma 2.3. Let CubeSimp K be the direct product of the lattices Simp KΨ , for Ψ ∈ ConM K: CubeSimp K = (Simp KΨ | Ψ ∈ ConM K). If the function Simp is understood, we write Cube K for CubeSimp K. We regard Simp KΨ , for Ψ ∈ ConM K, an ideal of Cube K, as in Section 2.2. For a ∈ K, deﬁne Diag(a) ∈ Cube K as follows: Diag(a) = a/Ψ | Ψ ∈ ConM K. K has a natural (diagonal) embedding into Cube K by ψ : a → Diag(a),

for a ∈ K.

72

6. Cubic Extensions

Let Diag(K) be the sublattice { Diag(a) | a ∈ K } of Cube K, and for a congruence Θ of K, let Diag(Θ) denote the corresponding congruence of Diag(K). By identifying a with Diag(a), for a ∈ K, we can view Cube K as an extension of K; we call Cube K a cubic extension of K.

1

1

Φ

c d

c

a

N6

b

d

a

b 0

0

f

Simp N6

Figure 6.1: The lattices N6 and Simp N6 . Cubic extensions are hard to draw because they are direct products. Here is a small example. We take the lattice N6 ; see Figure 6.1. The lattice N6 is subdirectly irreducible; it has two meet-irreducible congruences, ω and Φ. Since N6 /Φ = C2 , it is simple, so we can choose Simp N6 /Φ = C2 . The other lattice quotient is N6 /ω = N6 and for this we choose a simple extension Simp N6 , by adding an element; see Figure 6.1. The lattice Cube N6 is shown in Figure 6.2 along with the embedding Diag of N6 into Cube N6 . The images of elements of N6 under Diag are black-ﬁlled.

Diag(1)

Cube(Φ) F D(d)

B

D(b) = Diag(b) D(c) = Diag(c) D(d) = Diag(d)

D(c) D(a)

D(b)

Diag(0) Figure 6.2: The embedding of N6 into Cube N6 . As an alternative, you may draw the chopped lattice M whose ideal lattice is Cube N6 , see Figure 6.3. Unfortunately, the embedding Diag is not easy to see with this representation.

6.2. The basic property

73

Figure 6.3: The chopped lattice M whose ideals represent Cube N6 .

6.2.

The basic property

The following crucial property of cubic extensions was stated in Gr¨ atzer and Schmidt [69]. Theorem 6.1. Let K be a ﬁnite lattice and let Cube K be a cubic extension of K. Then K (= Diag(K)) is congruence-reflecting in Cube K. Proof. For Ω ∈ Con K and Ψ ∈ ConM K, deﬁne the congruence Cube(Ω, Ψ) on the lattice Simp KΨ as follows: ω, if Ω ≤ Ψ, Cube(Ω, Ψ) = ι, if Ω Ψ; and deﬁne Cube(Ω) =

( Cube(Ω, Ψ) | Ψ ∈ ConM K ),

a congruence of the lattice Cube K. Observe that Cube(ωK ) = ωCube K and Cube(ιK ) = ιCube K , that is, Cube(ω) = ω and Cube(ι) = ι. We show that Diag(Ω) is the restriction of Cube(Ω) to K. First, assume that Diag(u) ≡ Diag(v) (Diag(Ω)) in Diag(K). Then u ≡ v (Ω) in K, by the deﬁnition of D. The congruence u ≡ v (Ω) implies that u ≡ v (Ψ), for all Ψ ∈ ConM K satisfying Ω ≤ Ψ, that is, u/Ψ = v/Ψ, for all Ψ ∈ ConM K satisfying Ω ≤ Ψ. This, in turn, can be written as u/Ψ ≡ v/Ψ

(Cube(Ω, Ψ)), for all Ψ ∈ ConM K satisfying Ω ≤ Ψ,

since, by deﬁnition, ΩΨ = ω, for Ω ≤ Ψ. Again, by deﬁnition, ΩΨ = ι, for Ω Ψ. Therefore, the congruence u/Ψ ≡ v/Ψ (Cube(Ω, Ψ)) always holds. We conclude that u/Ψ ≡ v/Ψ (Cube(Ω, Ψ)), for all Ψ ∈ ConM K. This congruence is equivalent to Diag(u) ≡ Diag(v) (Cube(Ω)) in Cube K, which was to be proved.

74

6. Cubic Extensions

Second, assume that Diag(u) ≡ Diag(v) (Cube(Ω)) in Cube K. Then u/Ψ ≡ v/Ψ (Cube(Ω, Ψ)), for all Ψ ∈ ConM K; in particular, for all Ψ ∈ ConM K satisfying Ω ≤ Ψ. Thus u/Ψ = v/Ψ, for all Ψ ∈ ConM K satisfying Ω ≤ Ψ, that is, u ≡ v (Ψ), for all Ψ ∈ ConM K satisfying Ω ≤ Ψ. Therefore, u ≡ v ( ( Ψ ∈ ConM K | Ω ≤ Ψ )). The lattice Con K is ﬁnite, so every congruence is a meet of meet-irreducible congruences; therefore, Ω = ( Ψ ∈ ConM K | Ω ≤ Ψ ), and so u ≡ v (Ω) in K, that is, Diag(u) ≡ Diag(v) (Diag(Ω)) in Diag(K). For Ω ∈ Con K, the set ΔΩ = { Ψ ∈ ConM K | Ω Ψ } is a down-set of ConM K, and every down-set of ConM K is of the form ΔΩ , for a unique Ω ∈ Con K. The down-set ΔΩ of ConM K describes Cube(Ω), and conversely. In the example of Section 6.1 (see Figures 6.1–6.3), the congruence lattice Con N6 = {ω, Φ, ι}. Since Cube(ω) = ω and Cube(ι) = ι, we only have to compute Cube(Φ). Clearly, Cube(Φ) = ω × ι (that is, Cube(Φ) = ωC2 × ιSimp N6 ); it splits Cube N6 into two parts as shown by the dashed line in Figure 6.2. We summarize the properties of the lattice Cube K: Theorem 6.2. Let K be a ﬁnite lattice with a cubic extension Cube K. Then (i) Cube K is ﬁnite. (ii) Choose an atom sΨ ∈ Simp KΨ , for each Ψ ∈ ConM K, and let B be the boolean ideal of Cube K generated by these atoms. Then B is a congruence-determining ideal of Cube K. (iii) There are one-to-one correspondences among the subsets of ConM K, sets of atoms of B, and the congruences Ω of Cube K; the subset of ConM K corresponding to the congruence Ω of Cube K is ΔΩ = { Ψ ∈ ConM K | Ω Ψ }, which, in turn corresponds to the set SΩ = { sΨ | sΨ ≡ 0 (Ω) } of atoms of B. Hence, the congruence lattice of Cube K is a ﬁnite boolean lattice.

6.2. The basic property

75

(iv) Every congruence Ω of K has an extension Cube(Ω) to a congruence of Cube K corresponding to the down-set ΔΩ of ConM K. In the example of Section 6.1, for B we can choose the shaded ideal of Figure 6.2. Corollary 6.3. Choose a dual atom tΨ ∈ Simp KΨ , for each Ψ ∈ ConM K, and deﬁne tΨ as the element of Cube K whose Ψ-component is tΨ and all other components are 1. The element tΨ is a dual atom of Cube K. Let F be the boolean ﬁlter of Cube K generated by these dual atoms. Then F is a congruence-determining ﬁlter of Cube K. There are one-to-one correspondences among the subsets of ConM K, sets of dual atoms of F , and the congruences Ω of Cube K; the subset of ConM K corresponding to the congruence Ω of Cube K is ΔΩ = { Ψ ∈ ConM K | Ω Ψ }, which, in turn, corresponds to the set TΩ = { sΨ | sΨ ≡ 1 (Ω) } of dual atoms of F . Also, |F | = |B|. In the example of Section 6.1, for F we can choose the shaded ﬁlter of Figure 6.2. So a cubic extension Cube K of K (i) has a “cubic” congruence lattice (the boolean lattice Bn , an “n-dimensional cube”), and (ii) K and its cubic extension Cube K have the same number of meet-irreducible congruences, and (iii) the congruences Ω of K extend to Cube K (but, as a rule, the cubic extension has many more congruences than the Cube(Ω)-s). See Sections 12.2 and 13.2 for other small examples, in particular; see Figures 12.3 and 13.2.

Part III

Representation Theorems

7 The Dilworth Theorem

In this book we discuss a subﬁeld of Lattice Theory that started with the following result—a converse of the Funayama-Nakayama [21] result, Theorem 3.3 (page 37). Theorem 7.1 (Dilworth Theorem). Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnite lattice L. Our presentation is based on Gr¨ atzer and Schmidt [57], where the ﬁrst proof appeared. In his book (Crawley and Dilworth [10]), Dilworth reproduces the proof from [57]. It is clear from his recollections in [9] that his thinking was very close to ours. In this chapter we follow Gr¨ atzer and Lakser [34] (published in [26]), and prove this result based on the discussion of chopped lattices in Chapter 4, a simpler proof than the one in [57]. We will also prove that L can be constructed as a sectionally complemented lattice, as stated in [57].

7.1.

The representation theorem

By Theorem 4.6, to prove the Dilworth Theorem, it is suﬃcient to verify the following: Theorem 7.2. Let D be a ﬁnite distributive lattice. Then there exists a chopped lattice M such that Con M is isomorphic to D. Using the equivalence of nontrivial ﬁnite distributive lattices and ﬁnite orders (see Section 2.5.2) and using the notation ConJ M (see Section 3.2) for the order of join-irreducible congruences, we can rephrase Theorem 7.2 as follows:

80

7. The Dilworth Theorem

Theorem 7.3. Let P be a ﬁnite order. Then there exists a chopped lattice M such that ConJ M is isomorphic to P . We are going to prove the Dilworth Theorem in this form.

7.2.

Proof-by-Picture

The basic gadget for the construction is the lattice N6 = N (p, q) of Figure 7.1. The lattice N (p, q) has three congruence relations, namely, ω, ι, and Θ, where Θ is the congruence relation with congruence classes {0, q1 , q2 , q} and {p1 , p(q)}, indicated by the dashed line. Thus con(p1 , 0) = ι. In other words, p1 ≡ 0 “implies” that q1 ≡ 0, but q1 ≡ 0 “does not imply” that p1 ≡ 0. We will use the “gadget” N6 = N (p, q) to achieve such congruence-forcing.

p(q) q p1

q2

q1 0

Figure 7.1: The lattice N6 = N (p, q) and the congruence Θ. To convey the idea how to prove Theorem 7.3, we present three small examples in which we construct the chopped lattice M from copies of N (p, q). Example 1: The three-element chain. Let P = {a, b, c} with c ≺ b ≺ a. We take two copies of the gadget, N (a, b) and N (b, c); they share the ideal I = {0, b1 }; see Figure 7.2. So we can merge them (in the sense of Section 4.1) and form the chopped lattice M = Merge(N (a, b), N (b, c)) as shown in Figure 7.2. The congruences of M are easy to ﬁnd. The isomorphism P ∼ = ConJ M is given by x1 → con(0, x), for x ∈ P . The congruences of M can be described by a compatible congruence vector Θa,b , Θb,c (see Section 4.3), where Θa,b is a congruence of the lattice N (a, b) and Θb,c is a congruence of the lattice N (b, c), subject to the condition that Θa,b and Θb,c agree on I. Looking at Figure 7.1, we see that if the shared congruence on I is ω (= ωI ), then we must have Θa,b = ω (= ωN (a,b) ) and Θb,c = ω (= ωN (b,c) ) or Θb,c = Θ on N (b, c). If the shared congruence on I is ι (= ιI ), then we must have Θa,b = Θ or Θa,b = ι (= ιN (a,b) ) on N (a, b)

7.2. Proof-by-Picture

b(c)

a(b) b

N (a, b)

N (b, c)

b2

a1

I

c2

I 0

a(b)

P

b c

N (a, b) a1

b2

c

c1

b1 b1 0

a

81

b(c) b

N (b, c) b1

c

c2

0

c1

M

Figure 7.2: The chopped lattice M for P = C3 . and Θb,c = ι (= ιN (b,c) ) on N (b, c). So there are three congruences distinct from ω: ω, Θ, Θ, ι, ι, ι. Thus all the join-irreducible congruences form the three-element chain. Example 2: The three-element order PV of Figure 7.3. (We call PV the “order V .”) We take two copies of the gadget, N (b, a) and N (c, a); they share the ideal J = {0, a1 , a2 , a}; we merge them to form the chopped lattice MV = Merge(N (b, a), N (c, a)), see Figure 7.3. Again, the isomorphism PV ∼ = ConJ MV is given by x1 → con(0, x), for x ∈ PV . Example 3: The three-element order PH of Figure 7.4. (We call PH the “order hat.”) We take two copies of the gadget, N (a, b) and N (a, c); they share the ideal J = {0, a1 }; we merge them to form the chopped lattice MV = Merge(N (a, b), N (a, c)), see Figure 7.4. Again, the isomorphism PH ∼ = ConJ MH is given by x1 → con(0, x), for x ∈ PV . The reader should now be able to picture the general proof: instead of the few atoms in these examples, we start with enough atoms to reﬂect the

82

7. The Dilworth Theorem

b(a)

c(a) a

N (b, a) b

c

PV

a1

b1

a

N (c, a) c1

a2

MV

0

Figure 7.3: The chopped lattice for the order V .

a(b)

a(c) c

b N (a, b)

PH b

a

b1 c

N (a, c)

b2

a

c1

0

c2

MH

Figure 7.4: The chopped lattice for the order hat. structure of P , see Figure 7.5. Whenever b ≺ a in P , we build a copy of N (a, b), see Figure 7.6.

7.3.

Computing

For a ﬁnite order P , let Max be the set of maximal elements in P . We form the set M0 = {0} ∪ { p1 | p ∈ Max } ∪ ( {a1 , a2 } | a ∈ P − Max ) consisting of 0, the maximal elements of P indexed by 1, and two copies of the nonmaximal elements of P , indexed by 1 and 2. We make M0 a meet-semilattice by deﬁning inf{x, y} = 0 if x = y, as illustrated in Figure 7.5. Note that x ≡ y (Θ) and x = y imply that x ≡ 0 (Θ) and y ≡ 0 (Θ) in M0 ; therefore, the congruence relations of M0 are in one-to-one correspondence with subsets of P . Thus Con M0 is a boolean lattice whose atoms are associated with atoms of M0 ; the congruence Φx associated with the atom x has only one nontrivial block {0, x}. We construct an extension M of M0 as follows:

7.4. Sectionally complemented lattices

...

q1

p1

a1

a2

b2

b1

83

...

0 Figure 7.5: The chopped lattice M0 .

c(d) ...

c1

a(b)

d2

d1

a2

a1

b2

b

b1

...

0 Figure 7.6: The chopped lattice M . The chopped lattice M consists of four kinds of elements: (i) the zero, 0; (ii) for all maximal elements p of P , the element p1 ; (iii) for any nonmaximal element p of P , three elements: p, p1 , p2 ; (iv) for each pair p, q ∈ P with p q, a new element, p(q). For p, q ∈ P with p q, we set N6 = N (p, q) = {0, p1 , q, q1 , q2 , p(q)}. For x, y ∈ M , let us deﬁne x ≤ y to mean that, for some p, q ∈ P with p q, we have x, y ∈ N (p, q) and x ≤ y in the lattice N (p, q). It is easily seen that x ≤ y does not depend on the choice of p and q, and that ≤ is an ordering. Since, under this ordering, all N (p, q) and N (p, q) ∩ N (p , q ) (p q and p q in P ) are lattices and x, y ∈ M , x ∈ N (p, q), and y ≤ x imply that y ∈ N (p, q), we conclude that M is a chopped lattice; in fact, it is a union of the ideals N (p, q) with p q in P , and two such distinct ideals intersect in a one-, two-, or four-element ideal. Since the chopped lattice M is atomistic, Corollary 3.8 applies. If pi ⇒ qj in M , for p, q ∈ P and i, j ∈ {1, 2}, then p ≥ q in P , and conversely. So the equivalence classes of the atoms under the preordering ⇒ form an order isomorphic to Down P . This completes the veriﬁcation that ConJ M ∼ = P, and therefore, of Theorem 7.3.

7.4.

Sectionally complemented lattices

Let M be the chopped lattice described in Section 7.3. Since all N (p, q)-s are sectionally complemented, so is M . In Gr¨ atzer and Schmidt [57] the following result is proved:

84

7. The Dilworth Theorem

Theorem 7.4. L = Id M is sectionally complemented. Combined with Theorem 7.3, this yields the main result of this chapter: Theorem 7.5. Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnite sectionally complemented lattice L. To prove Theorem 7.4, we will view L = Id M as a closure system. For an ideal U of M , let Atom(U ) be the set of atoms of M in U . Note that the atoms of M are the {pi }, where p ∈ P and i ∈ {1, 2}. We start with the following two trivial statements (to simplify the notation, we compute with the indices modulo 2): Lemma 7.6. For a set A ⊆ Atom(M ), there is an ideal U with Atom(U ) = A iff A satisﬁes the condition: (Cl)

For p q in P if p1 , qi ∈ A, then qi+1 ∈ A.

Let us call a subset A of Atom(M ) closed if it satisﬁes (Cl). It is obvious that every subset A of Atom(M ) has a closure A. Lemma 7.7. The assignment I → Atom(I) is a bijection between the ideals of M and closed subsets of Atom(M ), and Atom(I ∧ J) = Atom(I) ∩ Atom(J), Atom(I ∨ J) = Atom(I) ∪ Atom(J), for I, J ∈ Id M . The inverse map assigns to a closed set X of atoms, the ideal id(X) of M generated by X. Lemma 7.7 allows us to regard L as the lattice of closed sets in Atom(M ), so I ∈ L will mean that I is a closed subset of Atom(M ). Thus I ∧ J = I ∩ J and I ∨ J = I ∪ J, for I, J ∈ L. Let I ⊆ J ∈ L. Let us say, that q ∈ P splits over I, J if there exists a p q in P , with p1 , qi ∈ J − I and qi+1 ∈ I. If there is a q ∈ P that splits over I, J, then J − I is not closed. Let X = X(I, J) be the set of all elements qi in J − I such that q splits over I, J. Let S(I, J) = (J − I) − X, that is, S(I, J) is the set of all elements qi in J − I such that q does not split over I, J. Lemma 7.8. S(I, J) ∈ L. Proof. We have to prove that S = S(I, J) is closed. Let u v in P , u1 ∈ S, and vi ∈ S. Since, by the deﬁnition of S, the element v does not split over I, J and u1 , vi ∈ J − I, it follows that vi+1 ∈ / I. Since u1 ∈ J and vi ∈ J and J is closed, we obtain that vi+1 ∈ J. Thus vi+1 ∈ J − I. Since v does not split over I, J, we get that vi+1 ∈ S by the deﬁnition of S. Thus S is closed.

7.5. Discussion

85

We claim that S = S(I, J) is the sectional complement of I in J. By Lemma 7.7, we have to prove that (1)

I ∩ S = ∅,

(2)

I ∪ S = J. (1) is obvious from the deﬁnition of S. Since I ⊆ J and S ⊆ J, to verify (2), it is suﬃcient to show that I ∪ S ⊇ J.

(3)

Assume, to the contrary, that there is a q ∈ P and i ∈ {1, 2} such that qi ∈ J − I ∪ S.

(4)

We can choose q so that it is maximal with respect to this property, that is, if p > q and pj ∈ J for some j ∈ {1, 2}, then pj ∈ I ∪ S. It follows from (4) that qi ∈ J − (I ∪ S) = X. So, by the deﬁnition of X, there is p q in P with p1 ∈ J − I and qi+1 ∈ I. Since p1 ∈ J and p q, by the maximality of q, we have that p1 ∈ I ∪ S. Also, qi+1 ∈ I ⊆ I ∪ S. So qi ∈ I ∪ S by the deﬁnition of closure, contradicting (4). This completes the proof of the claim and of Theorem 7.4

7.5.

Discussion

The Dilworth Theorem is the foundation of all the results in Parts III and IV. Moreover, the results of Part III derive from the results of Parts IV via the Dilworth Theorem. For instance, the result of Chapter 13 is: Every ﬁnite lattice K has a congruence-preserving embedding into a ﬁnite semimodular lattice L. Combining this with the Dilworth Theorem, we obtain the main result of Chapter 9: Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnite semimodular lattice L. Of course, this representation theorem is a relatively easy result compared to the congruence-preserving extension result. Nevertheless, the congruence-preserving extension theorem does not directly imply the representation theorem. (Note, however, that the proof of the result of Chapter 13 utilizes the result of Chapter 9.) An addendum The lattice L we construct for Theorem 7.1 has a very interesting congruence structure: every join-irreducible congruence Φ is of the form con(0, pΦ ), where pΦ is an atom and the pΦ -s generate a boolean ideal. We state this formally. Theorem 7.9 (Full Dilworth Theorem). Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnite sectionally complemented lattice L with the following property:

86

7. The Dilworth Theorem

(B) There is a boolean ideal B in L such that there is a bijection Φ → pΦ between ConJ L and the atoms of B satisfying Φ = con(0, pΦ ). There is a bijection between the congruences of L and “down-sets” of the atoms of B, that is, sets H of atoms satisfying the condition: pΦ ∈ H and Ψ ≤ Φ imply that pΨ ∈ H, for Φ, Ψ ∈ ConJ L. Proof. Let P = J(D) and take the sublattice B generated by { p1 | p ∈ P }. It is easy to see that B is an ideal. The statement about the congruences is implicit in the discussion of Section 7.3. The ideal B is, of course, a congruence-determining sublattice of L. Sectionally complemented chopped lattices The 1960 result of Gr¨ atzer and Schmidt [57]: Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnite sectionally complemented lattice L constructs a ﬁnite sectionally complemented chopped lattice M representing D, and proves that the ideal lattice of M is again sectionally complemented. We have tried with Schmidt for decades to prove a more general result of this type. It was only quite recently that in Gr¨ atzer, Lakser, and Roddy [44] an example of a ﬁnite sectionally complemented chopped lattice was found whose ideal lattice is not sectionally complemented; see Theorem 4.9 and Figure 4.2. Problem 7.1. Let M be a ﬁnite sectionally complemented chopped lattice. Find reasonable suﬃcient conditions under which Id M is sectionally complemented. The reader is referred to Gr¨ atzer and Lakser [41] and [42], Gr¨ atzer, Lakser, and Roddy [44], and Gr¨ atzer and Roddy [55], for relevant results. There are a number of problems that arise from Problem 7.1. We will describe a few; the reader should be able to easily add another dozen. As in Gr¨ atzer and Schmidt [66] and [63], we generalize chopped lattices from the ﬁnite to the general case. Let M be a meet-semilattice satisfying the following condition: (Ch) sup{a, b} exists for any a, b ∈ M having a common upper bound in M . We deﬁne in M : a ∨ b = sup{a, b}, whenever sup{a, b} exists in M . This makes M into a partial lattice called a chopped lattice. For a ﬁnite M , this is equivalent to the deﬁnition in Section 4.1. We deﬁne ideals, congruences, and merging in the general case as we did in the ﬁnite case. Unfortunately, the fundamental result, Theorem 4.6 of Gr¨ atzer and Lakser [34], fails for the general case.

7.5. Discussion

87

Problem 7.2. Investigate generalizations of Theorem 4.6 to the inﬁnite case. See Gr¨ atzer and Schmidt [63] for some related results. As we note in Section 4.2, every ﬁnite chopped lattice M decomposes into lattices: M = ( id(m) | m ∈ Max ). Problem 7.3. Can chopped lattices, in general, be usefully decomposed into lattices? Could this be utilized in Problem 7.2 by assuming that the chopped lattices decompose into ﬁnitely many lattices or into lattices with nice properties? Theorem 7.4 is interesting because the property of being sectionally complemented is not inherited, in general, when passing from a chopped lattice M to Id M . Let SecComp denote the class of sectionally complemented lattices and let SemiMod denote the class of semimodular lattices. For a class of lattices K, let Chop K denote the class of chopped lattices M with the property that id(m) ∈ K, for all m ∈ M . So Chop SecComp is what we call the class of sectionally complemented chopped lattices. Similarly, we could look at Chop SemiMod, and call its members semimodular. Problem 7.4. When is the ideal lattice of a ﬁnite semimodular chopped lattice again semimodular? Let us consider the following property of a class K of lattices: If M ∈ Chop K, then Id M ∈ K. Let us call such a class Chop -Id closed. Problem 7.5. Are there any nontrivial Chop -Id closed varieties? The mcr function For a natural number n and a class V of lattices, deﬁne mcr(n, V) (minimal congruence representation) as the smallest integer such that, for any distributive lattice D with n join-irreducible elements, there exists a ﬁnite lattice L ∈ V satisfying Con L ∼ = D and |L| ≤ mcr(n, V). We will investigate mcr(n, L) = mcr(n) in the next chapter. In this chapter we proved that mcr(n, SecComp) = O(22n ). Problem 7.6. Is mcr(n, SecComp) = O(22n ) the best possible result? For any class S of lattices if the Dilworth Theorem holds for S, then theoretically, the function mcr(n, S) exists, although it may be diﬃcult to compute. Of course, for any class S of lattices for which the Dilworth Theorem holds, we can raise the question what is mcr(n, S), and chances are that we get an interesting problem. In many cases, however, the Dilworth Theorem fails for S. Here are some nontrivial examples.

88

7. The Dilworth Theorem

(i) The congruence lattice of a ﬁnite relatively complemented lattice is boolean. So SecComp cannot be narrowed to the class of relatively complemented lattices. (ii) The class S = SecComp ∩ DuallySecComp would also be a logical candidate. Note, however, that by Theorem 11 of Gr¨ atzer and Schmidt [57] and Lemma 4.16 of M. F. Janowitz [94], every ﬁnite lattice in SecComp ∩ DuallySecComp has a boolean congruence lattice (see also [26], Theorem II.4.9). (iii) Similarly, one can ask whether SecComp can be narrowed to the class of semimodular sectionally complemented lattices. The discussion in Section IV.3 of [26] (in particular, the top paragraph of p. 240) shows that this cannot be done either. Every ﬁnite, semimodular, sectionally complemented lattice has a boolean congruence lattice. Interestingly, if a meaningfully deﬁned class S does not work, then as a rule, the congruence lattices of the ﬁnite lattices in the class are boolean. This suggest that to test a class S, one has to ﬁnd a ﬁnite lattice in S (the “gadget”) that represents the three-element chain as a congruence lattice. Problem 7.7. Is there a natural subclass S of SecComp for which Theorem 7.5 holds, that is, every ﬁnite distributive lattice D can be represented as the congruence lattice of a lattice L ∈ S.

Congruence class sizes Spectra The basic gadget in this chapter is the lattice N6 ; Figure 7.1 shows N6 and its only nontrivial congruence Θ. We can associate with the congruence Θ the pair 4, 2 measuring the size of the two congruence classes. Which pairs t1 , t2 can substitute for 4, 2? In other words, for which pairs of integers t1 , t2 is there a ﬁnite lattice L such that (1) L is sectionally complemented; (2) L has exactly one nontrivial congruence Θ; (3) Θ has exactly two congruence classes: a prime ideal P and a prime ﬁlter Q satisfying that |P | = t1 and |Q| = t2 . This question is answered as follows in Gr¨atzer and Schmidt [77]: Theorem 7.10. Let t1 , t2 be a pair of natural numbers. Then there is a ﬁnite lattice L with properties (1)–(3) iff t1 , t2 satisﬁes the following three conditions:

7.5. Discussion

89

(P1 ) 2 ≤ t1 and t1 = 3. (P2 ) 2 ≤ t2 and t2 = 3. (P3 ) t1 > t2 . What can we say about the cardinalities of the congruence classes of a nontrivial congruence in a ﬁnite sectionally complemented lattice? Let L be a ﬁnite lattice, and let Θ be a congruence of L. We denote by Spec Θ the spectrum of Θ, that is, the family of cardinalities of the congruence classes of Θ. In Gr¨ atzer and Schmidt [77], spectra are characterized for ﬁnite sectionally complemented lattices. Theorem 7.11. Let S = (mj | j < n) be a family of natural numbers, n ≥ 1. Then there is a nontrivial ﬁnite sectionally complemented lattice L and a nontrivial congruence Θ of L such that S is the spectrum of Θ iff S satisﬁes the following conditions: (S1 ) 2 ≤ n and n = 3. (S2 ) 2 ≤ mj and mj = 3, for all j < n. Corollary 7.12. Let S = (mj | j < n) be a family of natural numbers, n > 1. Then there is a nontrivial ﬁnite sectionally complemented lattice L with a unique nontrivial congruence Θ of L such that S is the spectrum of Θ iff S satisﬁes (S1 ) and (S2 ), and additionally: (S3 ) S is not constant, that is, there are j, j < n satisfying that mj = mj . (S4 ) n = 4. We only know that the congruence lattice Con L of the lattice L we construct in Theorem 7.11 has three or more elements. Can we prescribe its structure? Problem 7.8. Let D be a ﬁnite distributive lattice. Can we construct the lattice L of Theorem 7.11 that represents the given S = (mj | j < n) as the spectrum of a nontrivial congruence Θ so that L also satisﬁes D ∼ = Con L? The following form is even harder: Problem 7.9. Let D be a ﬁnite distributive lattice, and let a ∈ D − {0, 1}. Can we construct the lattice L of Theorem 7.11 that represents the given S = (mj | j < n) as the spectrum of a nontrivial congruence Θ so that L also satisﬁes D ∼ = Con L and under this isomorphism the element a of D maps to the congruence Θ of L? Note that the Corollary 7.12 solves this problem for D = C3 , the threeelement chain.

90

7. The Dilworth Theorem

Valuation There is a more sophisticated way of looking at spectra. Let K be a ﬁnite sectionally complemented lattice. Let us represent K in the form L/Θ, where L is a ﬁnite sectionally complemented lattice and Θ is a congruence of L. Then there is a natural map v : K → N (where N is the set of natural numbers) deﬁned as follows: Let a ∈ K; then a is represented by a congruence class A of Θ, so we can deﬁne v(a) = |A|. We call v a valuation on K. Theorem 7.13 (Gr¨ atzer and Schmidt [77]). Let K be a nontrivial ﬁnite sectionally complemented lattice, and let v : K → N. Then there exists a ﬁnite sectionally complemented lattice L and a nontrivial congruence Θ of L, such that there is an isomorphism ϕ : K → L/Θ satisfying v(a) = |aϕ|,

for all a ∈ K,

iff v satisﬁes the following conditions: (V1 ) v is antitone. (V2 ) 2 ≤ v(a) and v(a) = 3, for all a ∈ K. Corollary 7.14. Let K be a nontrivial ﬁnite sectionally complemented lattice, and let v : K → N. Then there exists a ﬁnite sectionally complemented lattice L and a unique nontrivial congruence Θ of L, such that there is an isomorphism ϕ : K → L/Θ satisfying v(a) = |aϕ|,

for all a ∈ K,

iff v satisﬁes the conditions (V1 ) and (V2 ), and additionally, v satisﬁes the following two conditions: (V3 ) v is not a constant function. (V4 ) K is simple. We can state Problems 7.8 and 7.9 also for valuations. Problem 7.10. Let D be a ﬁnite distributive lattice. Can we prove Theorem 7.13 with the additional condition: Con K ∼ = D? Problem 7.11. Let D be a ﬁnite distributive lattice, and let a ∈ D − {0, 1}. Can we prove Theorem 7.13 with Con K ∼ = D so that under this isomorphism a maps to Θ? Note that the Corollary 7.14 solves this problem for D = C3 , the threeelement chain. Here is a congruence-preserving extension variant of the valuation problem:

7.5. Discussion

91

Problem 7.12. Let L be a ﬁnite lattice and let Θ be a nontrivial congruence of L with spectrum v on K = L/Θ. Let v : K → N satisfy (V1 ) and (V2 ). If v ≤ v (that is, v(a) ≤ v (a), for all a ∈ K), then does there exist a ﬁnite congruence-preserving extension L of L such that the spectrum on L /Θ is v ?

8 Minimal Representations

8.1.

The results

In the proof of the Dilworth Theorem (Theorem 7.1), we construct—for a distributive lattice D with n ≥ 1 join-irreducible elements—a lattice L satisfying Con L ∼ = D. The size of this lattice is O(22n ). Can we do better? Using the notation of Section 7.5, what is mcr(n)? O(22n ) was improved to O(n3 ) in Gr¨ atzer and Lakser [35]—a substantial improvement from exponential to polynomial size—where it was conjectured that O(n3 ) can be improved to O(n2 ) and that O(n2 ) is best possible. Indeed, O(n2 ) is possible, as proved in Gr¨ atzer, Lakser, and Schmidt [45]. Theorem 8.1. Let D be a ﬁnite distributive lattice with n ≥ 1 join-irreducible elements. Then there exists a lattice L of O(n2 ) elements with Con L ∼ = D. In fact, there is such a planar lattice L. Or, equivalently, Theorem 8.1 . Let P be a ﬁnite order with n elements. Then there exists a lattice L of O(n2 ) elements with ConJ L ∼ = P . In fact, there is such a planar lattice L. The second part of the conjecture was veriﬁed in Gr¨ atzer, Rival, and Zaguia [54]. Theorem 8.2. Let α be a real number satisfying the following condition: Every distributive lattice D with n join-irreducible elements can be represented as the congruence lattice of a lattice L with O(nα ) elements. Then α ≥ 2.

94

8. Minimal Representations

p3 p3 p2 p2 p1 p2 p1 p1

p3

P

C

Figure 8.1: P and C.

8.2.

Proof-by-Picture for minimal construction

We present now the Proof-by-Picture of the minimal construction (Theorem 8.1 ) with the order P (the hat) of Figure 8.1. We form the chain C from P , as shown by the same diagram. The chain C is of length 2|P | = 6, it is colored (in the sense of Section 3.2) with elements of P as illustrated. We form C 2 , and color the prime intervals in the obvious way: a prime interval of C 2 is of the form [a, c, b, c] and [c, a, c, b], where a ≺ b in C, and we color both intervals the color of [a, b] in C. Figure 8.2 shows the two gadgets (colored lattices) we use: M3 and N5,5 . We proceed as follows: To construct L, we take C 2 with the coloring we have just introduced. If both lower edges of a covering square in C 2 have the same

N5,5 M3 q p

p

q

Figure 8.2: The two gadgets (p < q).

p

8.3. The formal construction

95

color, we add an element to make it a covering M3 , the ﬁrst gadget. If in C 2 we have a covering C3 × C2 , where the C2 is colored by p, the C3 is colored by q twice, where p < q in P , then we add an element to make it an N5,5 , the second gadget. What do these gadget do? If p and q are any two prime intervals of M3 , then p ⇔ q (using the notation of Section 3.2). If p and q are any two prime intervals of N5,5 of color p and q with p < q, then q ⇒ p (but p ⇒ q fails). The lattice L we obtain is shown in Figure 8.3. (The copies of N5,5 in the diagram are marked by black-ﬁlled elements.) Every prime interval of C 2 is projective to a prime interval in one of the two copies of C in C 2 . The ﬁrst gadget makes sure that a prime interval in the ﬁrst copy of C is projective to a prime interval of the same color in the second copy of C. The second gadget orders the prime intervals: if p is of color p and q is of color q, then p > q implies that p ⇒ q. Now Theorem 3.7 shows that the equivalence class of prime intervals of L form an order isomorphic to P , that is, ConJ L ∼ = P. In particular, we make both copies of C congruence-determining sublattices of L. What is the size of L? The size of C 2 is (2n+1)2 . For each color p ∈ P , we add the ﬁrst gadget four times, adding 4n elements. For all p < q ∈ P , there is only one pair of adjacent prime intervals of color q in the ﬁrst copy of C, and these can be paired with two prime intervals of color p in the second copy of C, yielding two more elements. Since there are at most n2 /2 such pairs, we get at most n2 new elements. So |L| ≤ (2n + 1)2 + 4n + n2 , so |L| is obviously O(n2 ). This completes the Proof-by-Picture.

8.3.

The formal construction

We now formally prove Theorem 8.1 . Let P be a ﬁnite order of n elements. Let P0 = {a0 , a1 , . . . , ak−1 } be the set of nonminimal elements of P . If k = 0, then P is an antichain, and we can choose L as a chain of length n. Henceforth, we assume that k > 0. Let C0 be a chain of length 2k. We color the prime intervals of C0 as follows: we color the lowermost two prime intervals of C0 with a0 , the next two with a1 , and so on. Thus, for each a ∈ P0 , there are in C0 two prime intervals of color a (and they are successive). So, for each a ∈ P0 , there is a unique subchain ab ≺ am ≺ at such that the prime intervals [ab , am ] and [am , at ] have color a, and no other prime interval of C0 has color a. Observe that, for each a distinct from a0 , there is a c ∈ P0 with ab = ct , and, similarly, for each a distinct from ak−1 , there is a c ∈ P0 with at = cb . The elements am , however, are labeled uniquely.

96

8. Minimal Representations

p3

p3 p3

C

p3 p2

p2 p2

p2 p1

C

p1 p1

p1

Figure 8.3: The minimal lattice L. Let C1 be a chain of length n = |P |. We color the prime intervals of C1 by an arbitrary bijection with P . Thus, for each a ∈ P , there is in C1 exactly one prime interval of color a; we denote it by [ao , ai ]. We set L0 = C0 × C1 . We will regard C0 and C1 as sublattices of L0 in the usual manner. As discussed in Section 3.2, we can preorder the prime intervals of L0 by ⇒ and we can choose the prime intervals in C0 ∪ C1 as representatives of the equivalence classes. Hence, the order of join-irreducible congruences of L0 is an antichain of cardinality 2k + n. Note that |L0 | = (2k + 1)(n + 1). We next extend the lattice L0 to a lattice L1 . For each a ∈ P0 , we adjoin two new elements m0 (a) and m1 (a) to L0 ; we set ab , ao ≺ m0 (a) ≺ am , ai , am , ao ≺ m1 (a) ≺ at , ai . The resulting order L1 is a lattice, and, for each a ∈ P0 , the intervals [ab , ao , am , ai ] = {ab , ao , am , ao , m0 (a), ab , ai , am , ai }, [am , ao , at , ai ] = {am , ao , at , ao , m1 (a), am , ai , at , ai } are isomorphic to M3 . By adjoining these elements, we have made congruence equivalent any two prime intervals of L1 in C0 ∪ C1 of the same color. Therefore, the order of join-irreducible congruences of L1 is isomorphic to the antichain P .

8.4. Proof-by-Picture for minimality

97

Note that both C0 and C1 are congruence-preserving sublattices of L1 . Observe that |L1 | = (2k + 1)(n + 1) + 2k. We ﬁnally further extend L1 so as to induce the correct order on the join-irreducible congruences. For each pair a c in P (whereby a ∈ P0 , necessarily), we add a new element n(a, c) to L1 , setting am , co ≺ n(a, c) ≺ am , ci . The resulting order L is a lattice. In L, the interval [am , at ] is projective with [n(a, c), am , ci ], which in turn is projective with [co , ci ]. So the order of join-irreducible congruences of L is isomorphic to P . Note that L is a planar lattice, and that in going from L1 to L, we adjoin no more than kn elements. Thus |L| ≤ (2k + 1)(n + 1) + 2k + kn < 3(n + 1)2 .

8.4.

Proof-by-Picture for minimality

We present a Proof-by-Picture of Theorem 8.2 by contradiction. Let α < 2 be a real number such that every order P with n elements can be represented as ConJ L, for a lattice L with O(nα ) elements, that is, |L| ≤ Cnα , for some constant C. Let p = [a, b] and q = [c, d] be prime intervals of L such that con(p) con(q) in ConJ L. Then by Lemma 3.6, p ⇒ q, for instance, u

d

p = [a, b] ∼ [u1 , v1 ] [u2 , v2 ] [u3 , v3 ] ∼ q = [c, d], as illustrated in Figure 8.4. Note the elements hi ∈ [ui , vi ], i = 1, 2, 3, d

u

d

satisfying [u1 , h1 ] ∼ [u2 , v2 ], [h2 , v2 ] ∼ [u3 , v3 ], and [h3 , v3 ] ∼ q.

b p a

v1

v3

e2

h1

h3

con(p)

u1

v2 u3 h2 u2

d q c

h con(q) e1

e2 con(q) h con(p) e1

Figure 8.4: p ⇒ q illustrated for con(p) con(q) in ConJ L.

98

8. Minimal Representations

Since con(p) > con(q) and con(p) ≥ con(u1 , v1 ) ≥ con(u2 , v2 ) ≥ con(u3 , v3 ) ≥ con(q), we must have a > in the last formula; assume, for instance, that con(p) = con(u1 , v1 ) > con(u2 , v2 ). Then con(u1 , h1 ) = con(u2 , v2 ) < con(p), so con(h1 , v1 ) = con(p). Since q is collapsed by con(u1 , h1 ), by Lemma 3.5, there is a prime interval [e1 , x] in [u1 , h1 ] so that q is collapsed by con(e1 , x). Now con(p) > con(e1 , x) ≥ con(q), so by the assumption con(p) con(q), it follows that con(e1 , x) = con(q). So (with h = h1 and e2 = v1 ) we found a three-element chain e1 < h < e2 with con(e1 , h) = con(q) and con(h, e2 ) = con(p). Other choices lead to the same result, with maybe con(q) on the top. The rest is simple combinatorics. Let P2n = {k1 , . . . , kn , m1 , . . . , mn } (= Bn ) be the bipartite graph with minimal elements k1 , . . . , kn , maximal elements m1 , . . . , mn , so ki ≺ mj in P , for 1 ≤ i, j ≤ n, and {k1 , . . . , kn }, {m1 , . . . , mn } are antichains. Figure 8.5 shows B3 .

k1

k2

k3

m1

m2

m3

Figure 8.5: The order B3 . Let L2n be a lattice with ConJ L2n = Bn and satisfying |L2n | ≤ C(2n)α = C n , for some constant C and C = 4C. For 1 ≤ i, j ≤ n, we get in L2n a chain C(i, j): e1 (i, j) < h(i, j) < e2 (i, j) in L. Since there are n2 elements of the form h(i, j) and there are at most C nα elements in L2n , therefore, there is an element h occurring as the middle element in n2 /(C nα ) (that is, C1 n2−α ) of these chains. For h(i, j), either con(h(i, j), e1 (i, j)) or con(h(i, j), e2 (i, j)) 1 2−α is a minimal element of Bn . So there are 2C of these elements for which n these choices are consistent, say, all con(h(i, j), e1 (i, j)) are maximal. Let A be the set of these e1 (i, j)-s. α

...

x

...

A

h Figure 8.6: A conﬁguration too large in L. We get the situation depicted in Figure 8.6: for any x ∈ A, the congruence con(h, x) corresponds to a maximal element of Bn . Therefore, A is

8.5. Computing minimality

99

an antichain. Since a ﬁnite join of a subset of an antichain of join-irreducible elements cannot contain an element of the antichain not in the subset, we conclude that A under join generates 2|A| − 1 distinct elements. From 1 2−α |A| ≥ 8C n , it now follows that 2|A| − 1 = C1 2n

2−α

, 2−α

for some constant C1 . So we conclude that L2n has at least C1 2n elements, contradicting that L2n has at most C nα elements, completing the Proof-byPicture.

8.5.

Computing minimality

We start by formalizing the situation depicted in Figure 8.4. Lemma 8.3. Let L be a ﬁnite lattice, and let vi , ui ∈ L satisfy vi ≺ ui , for i = 1, 2. Let Φi = con(vi , ui ), for i = 1, 2. If Φ1 ≺ Φ2 in ConJ L, then there is a three-element chain {e1 , h, e2 } in L such that Φi = Θ(h, ei ), for i = 1, 2, and e1 < h < e2 or e2 < h < e1 . Proof. Since v1 ≡ u1 (con(v2 , u2 )) and v1 ≺ u1 , by Theorem 3.1, there is a sequence of congruence-perspectivities [v2 , u2 ] = [x1 , y1 ] [x2 , y2 ] [x3 , y3 ] . . . [xn , yn ] = [v1 , u1 ], for some natural number n > 1. Obviously, Φ2 = con(x1 , y1 ) = con(x2 , y2 ) ≥ con(x3 , y3 ) ≥ · · · ≥ con(xn , yn ) = Φ1 . Since Φ2 > Φ1 , there is a smallest i satisfying con(xi , yi ) < Φ2 ; obviously, 3 ≤ i ≤ n. Let i be odd, and let z = xi−1 ∨ yi . Then con(xi−1 , z) = con(xi , yi ) < Φ2 , but con(xi−1 , yi−1 ) = Φ2 , so con(z, yi−1 ) = Φ2 . Since u1 ≡ v1 (con(xi−1 , z)) and u1 ≺ v1 , there are u, v ∈ [xi−1 , z], such that v ≺ u and con(u1 , v1 ) ≤ con(u, v). So Φ1 = con(u1 , v1 ) ≤ con(u, v) < Φ2 , and con(u, v) is a join-irreducible congruence, hence by the assumption on Φ1 and Φ2 , it follows that Φ1 = con(u, v). Hence we can choose e2 = xi−1 , h = z, and e1 = yi−1 . If i is even, then we proceed dually. The second lemma deals with join-independence. A set A in a ﬁnite lattice L is join-independent if for any a ∈ A and subset A1 ⊆ A, the inequality a ≤ A1 implies that a ∈ A1 . Lemma 8.4. Let L be a ﬁnite lattice, let A ⊆ L, and let b ∈ L be a lower bound of A in L. If { con(b, x) | x ∈ A } is join-independent in Con L, then A is join-independent in L.

100

8. Minimal Representations

Proof. Indeed, if a ≤

A1 for some A1 ⊆ A, then con(b, a) ≤ { con(b, x) | x ∈ A1 },

a contradiction. Finally, observe that in a ﬁnite distributive lattice, a set A of join-irreducible elements is join-independent iﬀ the elements are pairwise incomparable. So we obtain the following. Corollary 8.5. Let L be a ﬁnite lattice, let A ⊆ L, and let b ∈ L be a lower bound of A in L. If { con(b, x) | x ∈ A } is a set of pairwise incomparable join-irreducible congruences, then A is join-independent in L. Now we prove Theorem 8.2 with the combinatorial argument of Section 8.4, bolstered by Corollary 8.5.

8.6.

Discussion

History In the early research, Dilworth, Gr¨ atzer, and Schmidt used only the N6 gadget for constructing lattices with given congruence lattices. Thus the size O(22n ). About 30 years later, I tackled the old problem of Birkhoﬀ [6]: characterizing congruence lattices of inﬁnitary algebras (see [27] and [28]); in fact, in the stronger form conjectured by R. Wille (see Reuter and Wille [103]): Every complete lattice can be represented as the lattice of complete congruences of a suitable complete lattice. The ﬁnite case was earlier done in Teo [115]. With Lakser (see [37]), we found a substantially simpler proof of my result, using colored chains; see also Gr¨ atzer and Schmidt [62]. Next year, we realized (see [38], result announced in Gr¨ atzer and Lakser [36]) that this technique can be utilized to prove that Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnite lattice L of size O(n3 ), where n is the number of join-irreducible elements of D. Soon thereafter, using similar techniques, Gr¨ atzer, Lakser, and Schmidt [45] found the O(n2 ) result. The third ﬁeld, not mentioned On page xiii of the Preface, we say that the papers ﬂowing from the Dilworth Theorem split into two ﬁelds: the ﬁnite and the inﬁnite case. Actually, there is also a third ﬁeld: complete congruence lattices of complete lattices. This ﬁeld started as described in the history section above. Here are some additional papers from this ﬁeld, in chronological order: Gr¨ atzer, Lakser, and Wolk [52], Freese, Gr¨atzer, and Schmidt [17], Gr¨ atzer and Lakser [39], Gr¨ atzer and Schmidt [59], [60], [64], [65], [67], [73]. The best result in this ﬁeld is from Gr¨ atzer and Schmidt [64].

8.6. Discussion

101

Theorem 8.6. Every complete lattice can be represented as the complete congruence lattice of a complete distributive lattice. The class D of distributive lattices is the minimal nontrivial variety, so this result cannot be improved by replacing D by a smaller variety. However, there are special classes of complete distributive lattices. The two best known inﬁnitary identities are the Join-Inﬁnite Distributive Identity: (JID) a ∧ X = ( a ∧ x | x ∈ X ), and its dual, the Meet-Inﬁnite Distributive Identity: (MID) a ∨ X = ( a ∨ x | x ∈ X ). We will denote by (JIDm ) the condition that (JID) holds for sets X satisfying |X| < m, where m is a regular cardinal with m > ℵ0 . We deﬁne (MIDm ) dually. The lattice we construct for Theorem 8.6 in Gr¨ atzer and Schmidt [64] atzer and Schmidt [67], we prove the fails both (JIDm ) and (MIDm ). In Gr¨ following result. Theorem 8.7. Let L be a complete lattice with more than two elements and with a meet-irreducible zero. Then L cannot be represented as the lattice of complete congruence relations of a complete distributive lattice K satisfying (JID) and (MID). So we can raise the following: Problem 8.1. Characterize the complete congruence lattices of complete distributive lattices satisfying (JID) and/or (MID). Or more generally: Problem 8.2. Characterize the lattices of m-complete congruences of m-complete distributive lattices satisfying (JIDm ) and/or (MIDm ). We can also ask: Problem 8.3. Characterize the lattices of complete (resp., m-complete) congruences of complete (resp., m-complete), weakly atomic, distributive lattices. Improved bounds A somewhat sharper form of Theorem 8.1 is the following: Theorem 8.8. For any integer n ≥ 2, 1 n2 < mcr(n) < 3(n + 1)2 . 16 log2 n

102

8. Minimal Representations

The upper bound was proved in Gr¨ atzer, Lakser, and Schmidt [45] and the lower bound in Gr¨ atzer and Wang [80]. Zhang [123] noticed that the proof of this inequality can be improved to obtain the following result: for n ≥ 64, 1 n2 < mcr(n). 64 (log2 n)2 Diﬀerent approaches to minimality A diﬀerent kind of lower bound is obtained in Freese [16]; it is shown that if ConJ L has e edges (e > 2), then e ≤ |L|. 2 log2 e Freese also proves that ConJ L can be computed in time O(|L|2 log2 |L|). Consider the optimal length of L. Schmidt [107] constructs a ﬁnite lattice L of length 5m, where m is the number of dual atoms of D (for ﬁnite chains, this was done in Berman [5]); Teo [115] proves that this result is best possible. Problem 8.4. In Theorem 8.2, is the construction “best” is some sharper sense? In other words, is there a function f (n) with the properties: (i) Theorem 8.1 holds with O(f (n)) in place of O(n2 ); (ii) f (n) < n2 ; (iii) n2 = O(f (n)) fails? Yet another approach to minimality starts with the elegant inequality for a ﬁnite lattice L: J(L) ≥ | ConJ L| of Freese, Jeˇzek, and Nation [18]. So if we deﬁne je(L) = J(L) − | ConJ L|, then je(L) ≥ 0 and je(L) is one measure of the eﬃciency of the representation of D = Con L as a congruence lattice. For a ﬁnite distributive lattice D, deﬁne JE(D) = min(je(L) | Con L ∼ = D), where L ranges over ﬁnite lattices. Then from this point of view, the best representation of a ﬁnite distributive lattice D as a congruence lattice of a ﬁnite lattice L is obtained when je(L) = JE(Con L). This is the approach we take in Gr¨ atzer and Wehrung [87]. In Gr¨ atzer and Wehrung [87], we compute the exact value of JE(D), yielding the inequality 2 0 ≤ JE(D) ≤ n, 3 and the constant 2/3 in this estimate is best possible.

8.6. Discussion

103

Problem 8.5. Determine the least constant k such that for every ﬁnite distributive lattice D, there exists a ﬁnite sectionally complemented lattice L such that Con L ∼ = D and J(L) ≤ k|J(D)|. By Gr¨ atzer and Schmidt [57], this constant k ≤ 2. The value of the constant deﬁned similarly for the class of atomistic lattices (or the class of all lattices as well) equals 5/3, by Corollary 5.4 of Gr¨ atzer and Wehrung [87]. Problem 8.6. Let V be a variety of lattices. If D is a ﬁnite distributive lattice representable by a ﬁnite lattice in V, compute the least possible value of J(L), for a ﬁnite lattice L in V such that Con L ∼ = D. A general lattice construction In this chapter, as well as some others, we use the simple trick of adding an element to a lattice to make a covering B2 into an M3 . We do the same to insert copies of N5,5 , and in Chapter 9, to insert copies of S8 . Gr¨ atzer and Kelly [33] discuss a general lattice construction containing these special cases.

9 Semimodular Lattices

9.1.

The representation theorem

Should we try to get a representation theorem for semimodular lattices? For a class to be a candidate, it would have to satisfy two criteria: First, there ought to be very many ﬁnite simple lattices in the class. For semimodular lattices, the ﬁnite partition lattices (Section 1.2.1) are all simple semimodular lattices by Ore [99], so we have lots of ﬁnite simple semimodular lattices. Second, there ought to be ﬁnite lattices in the class with a three-element congruence lattice. For semimodular lattices, the lattice S8 of Figure 9.1 is such a lattice.

p p p q

p p q

q

p

p

Figure 9.1: The semimodular gadget S8 , colored with p < q. These criteria satisﬁed, we looked for a representation theorem for semimodular lattices.

106

9. Semimodular Lattices

In fact, in Gr¨ atzer, Lakser, and Schmidt [48], we found a very good representation theorem for semimodular lattices, also obtaining—to our great surprise—planarity and a small size: Theorem 9.1. Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnite semimodular lattice L. In fact, if D has n ≥ 1 join-irreducible elements, then L can be constructed as a planar lattice of size O(n3 ). By Theorem 8.1, the optimal size for any lattice is O(n2 ); the size we obtain in this result seems very good.

9.2.

Proof-by-Picture

The proof of this result is very similar to the proof of Theorem 8.1. The gadget we use now is the semimodular (colored—in the sense of Section 3.2) lattice S8 of Figure 9.1.

a b

d

e

c Figure 9.2: The order P . To represent the order P of Figure 9.2 as ConJ L, for a ﬁnite planar semimodular lattice L, we start out—see Figure 9.3—by constructing the lattices

a q a

a

d

p

b

a

a

b b

d

b

Aa

b

Figure 9.3: The lattices Aa and Ab .

c

c

Ab

9.3. Construction and proof

107

Aa and Ab . We obtain Aa by gluing two copies of S8 together; and color it with {a, b, d} so that con(a) > con(b) is accomplished in the top S8 of Aa and con(a) > con(d) is accomplished in the bottom S8 of Aa . The lattice Ab is S8 colored by {b, c} so that con(b) > con(c) in Ab . Observe that the lattice Aa takes care of all a x orderings; in the example, there are only two. We could do three coverings by gluing three copies of S8 together, and so on. Form the glued sum S of Aa and Ab ; all the covers of P are taken care of in S. There is only one problem: S is not a colored lattice; in this example if p is a prime interval of color b in Sb (as in Figure 9.3) and q is a prime interval of color b in Sb (as in Figure 9.3), then in S we have con(p) ∧ con(q) = ω. Of course, we should have con(p) = con(q) = con(b). We accomplish this by extending S to the lattice L of Figure 9.4. In L, the black-ﬁlled elements form the sublattice S. As you see, we extend S by adding to it a distributive “grid.” The right corner is C52 colored by {a, b, c, d}; each of the four covering squares colored by the same color twice are made into a covering M3 . This makes the coloring behave properly in the right corner. In the rest of the lattice we do the same: we look for a covering square colored by the same color twice, and make it into a covering M3 . This makes L into a colored lattice: any two prime intervals of the same color generate the same congruence. For instance, if p and q are prime intervals as in the previous paragraph (see now Figure 9.4), then we ﬁnd in L the prime intervals r1 , r2 , r3 , r4 (see Figure 9.4), so that p r1 r2 r3 r4 q; therefore, conL (p) = conL (q). Finally, we remember the element e ∈ P . We add a “tail” to the lattice and color it e. It is easy to see that the resulting lattice L is planar and semimodular and that ConJ L is isomorphic to P ; the isomorphism is x → con(x), for x ∈ {a, b, c, d, e}.

9.3.

Construction and proof

We construct the semimodular lattice L of Theorem 9.1 in several steps. Take the lattice S8 with the notation of Figure 9.5. The lattice S8 has an ideal, I8 , and a ﬁlter, F8 , both isomorphic to C2 ; we will utilize these for repeated gluings—deﬁned in Section 2.4. The elements of I8 and F8 are black-ﬁlled in Figure 9.5. Let D be a ﬁnite distributive lattice, and let P = J(D) be the order of its join-irreducible elements, n = |P |. We enumerate p1 , p2 , . . . , pm the non-minimal elements of P . For every pi , i = 1, 2, . . . , m, let cov(pi ) = {pi1 , pi2 , . . . , piki }

108

9. Semimodular Lattices

d

r1 c p

b

b a r3

a c b

r2

a c b b q

c

b d d

r4

d

e

Figure 9.4: The lattice L. denote the set of all lower covers of pi in P ; since pi is non-minimal, it follows that ki > 0. Let r1 , r2 , . . . , rt enumerate all elements of P that are incomparable with all other elements. In the example of Section 9.2, we have, say, a = p1 , b = p2 , k1 = 2, k2 = 1, b = p11 , d = p12 , c = p21 , cov(a) = {b, d}, and cov(b) = {c}. Step 1: For every i, with 1 ≤ i ≤ m, we construct a lattice Ai with an

109

9.3. Construction and proof

i8 F8

a8 e8

c8 f8 d8

b8 I8 o8

Figure 9.5: The semimodular gadget S8 , with notation. ideal Ii and a ﬁlter Fi , where Ii is a chain of length 2(ki + · · · + km ) and Fi is a chain of length 2(ki+1 + · · · + km ). For the order P of Section 9.2, we construct the lattices A1 and A2 ; see Figure 9.6—the elements of the ideals I1 and I2 , and of the ﬁlters F1 and F2 are black-ﬁlled.

F1

F2 I1

I2

A1

A2

Figure 9.6: The lattices A1 and A2 Now we will use twice the construction, gluing k-times, described in Section 2.4. To form Ai , glue S8 to itself (ki − 1)-times with the ideal I8 and the ﬁlter F8 , to obtain the lattice A1i with a ﬁlter FA1i .

110

9. Semimodular Lattices

Now take B2 = {0, 0, 0, 1, 1, 0, 1, 1} with the ideal IB2 = {0, 0, 1, 0} and the ﬁlter FB2 = {0, 1, 1, 1}, and glue 2(ki+1 + · · · + km )-times B2 to A1i . The ideal Ii is generated by the element 0, 1 of the top B2 , while Fi is generated by the unit element of A1i . We deﬁne a coloring μi of Ai as follows. On any copy of S8 , [o8 , b8 ]μi = pi and on the j-th copy of S8 , [o8 , d8 ]μi = [d8 , c8 ]μi = pij ; on the ﬁrst two copies i+1 of B2 , [0, 1, 1, 1]μi = pi+1 1 , on the next two copies, [0, 1, 1, 1]μi = p2 , i+2 after ki+1 pairs, the next two satisﬁes [0, 1, 1, 1]μi = p1 , and so on. Lemma 9.2. μi is a coloring of Ai . The join-irreducible congruences of Ai are generated by prime intervals of Ii and by [o8 , b8 ] of the bottom S8 in Ai . If p and q are [o8 , b8 ] or a prime interval [o8 , d8 ] or [d8 , c8 ] of a copy of S8 in Ai , then Θ(p) ≥ Θ(q) iﬀ pμi ≥ qμi . In particular, con(o8 , b8 ) con(o8 , d8 ) in ConJ Ai , where o8 , b8 , d8 are in a copy of S8 in Ai . If p is a prime interval [0, 1, 1, 1] in a copy of B2 , then con(p) is incomparable to any con(q), where q is [o8 , b8 ] or a prime interval of Ii diﬀerent from p. Proof. This is trivial since every prime interval of S8 is projective to one of [o8 , b8 ], [o8 , d8 ], [d8 , c8 ]. Step 2: We deﬁne the lattice A by gluing together the (colored) lattices Ai , 1 ≤ i ≤ m. For the order P of Section 9.2, we construct the lattice A; see Figure 9.7. For 1 ≤ i ≤ m, we deﬁne, by induction, the lattice Ai , which contains Ai , and, therefore, Fi , as a ﬁlter. Let A1 = A1 . Assume that Ai with Fi as a ﬁlter has been deﬁned. Observe that both Fi and Ii+1 are chains of length 2(ki+1 + · · · + km ), and so they are isomorphic; in fact, this isomorphism preserves colors. We glue Ai to Ai+1 over Fi and Ii+1 to obtain Ai+1 . Deﬁne A = Am and IA = I1 (see Figure 9.7). Observe that μi on Fi agrees with μi+1 on Ii+1 ; therefore, the μi deﬁne a coloring μA of A, for 1 ≤ i ≤ m. Let FA be the ﬁlter of A generated by the element 0, 1 of the top B2 in A1 . FA is a chain of length m. The prime interval [o8 , b8 ] in the bottom S8 in Ai (1 ≤ i ≤ m) is projective to a unique prime interval p of FA ; deﬁne pμA = [o8 , b8 ]μA . Lemma 9.3. μA is a coloring of A. The join-irreducible congruences of A are generated by prime intervals of IA and FA . Let p and q be prime intervals in IA and FA .

9.3. Construction and proof

111

FA

IA

Figure 9.7: The lattice A. (i) If p and q are prime intervals of FA , then con(p) and con(q) are incomparable. (ii) If p is a prime interval of FA and q is a prime interval of IA , then con(p) and con(q) are comparable iﬀ p ⊆ Ai , for some 1 ≤ i ≤ m, q is perspective to some [o8 , d8 ] or [d8 , c8 ] in some S8 in Ai ; in which case, con(p) con(q) in ConJ A. (iii) If p and q are prime intervals of IA , then con(p) ≥ con(q) iﬀ p and q are perspective to prime intervals p and q in some Ai , respectively, for some 1 ≤ i ≤ m, and p and q are adjacent edges of some S8 in Ai , in which case, con(p) = con(q). Proof. This is obvious from Lemma 2.9. Observe that the congruence lattice of A is still quite diﬀerent from D in two ways, exactly as we discussed this for S in Section 9.2: the congruences that correspond to the ri -s are still missing; prime intervals in IA ∪ FA of the same color generate incomparable congruences with one exception: they are adjacent intervals in IA , perspective to the two prime intervals of some S8 in some Ai .

112

9. Semimodular Lattices

Step 3: We extend A to a lattice B with an ideal IB which is a chain and which has the property that every prime interval of B is projective to a prime interval of IB . This step is easy. We form the lattice FA2 with the ideal IFA2 = { x, 0FA | x ∈ FA }, where 0FA is the zero of FA . Let 1FA denote the unit element of FA and, for x ∈ FA , x < 1FA , let x∗ denote the cover of x in FA . For every x ∈ FA , x < 1FA , we add an element mx to FA2 so that the elements x, x, x, x∗ , x∗ , x, xm , x∗ , x∗ form a sublattice isomorphic to M3 with x, x as zero and x∗ , x∗ as unit. Let M be the resulting lattice. Obviously, M is a ﬁnite planar modular lattice whose congruence lattice is isomorphic to the congruence lattice of FA . Clearly, IFA2 is also an ideal of M ; we will denote it by IM . We glue A to M over FA and IM to obtain B. Let IB be deﬁned as the ideal generated by 0, 1FA . We deﬁne μB as an extension of μA ; every prime interval p of M is projective to exactly one prime interval p of IM , we deﬁne pμB = pμA . Lemma 9.4. μB is a coloring of B. The join-irreducible congruences of B are generated by prime intervals of IB . Let p and q be prime intervals in IB . (i) If p and q are prime intervals of M , then con(p) and con(q) are incomparable. (ii) If p is a prime interval of M and q is a prime interval of IA , then con(p) and con(q) are related exactly as conA (p) and conA (q) are related in A. (iii) If p and q are prime intervals of IA , then con(p) and con(q) are related exactly as conA (p) and conA (q) are related in A. Proof. This is obvious from the congruence structure of M . Step 4: We extend B to the lattice L of the Theorem 9.1, as sketched in Figure 9.8. This is also an easy step. We take a chain C of length n and we color C over P so that the coloring is a bijection. We form the lattice C × IB . For every pair of prime intervals, p = [a, b] of C and q = [c, d] of IB if p and q have the same color, then we add an element m(p, q) to C over P so that the elements a, c, b, c, a, d, m(p, q), b, d

9.3. Construction and proof

113

form a sublattice isomorphic to M3 . Let N denote the resulting lattice. N is obviously modular and planar. Set IN = { x, 0IB | x ∈ C }, FN = { 1C , x | x ∈ IB }, where 0IB is the zero of IB and 1C is the unit of C. Then IN is the ideal of N (isomorphic to C) and FN is a ﬁlter of N (isomorphic to IB ). Every prime interval of N is projective to a prime interval of IN , so we have a natural coloring μN on N . Note that this coloring agrees with the coloring μB on FN under the isomorphism with IB . We glue N to B over FN and IB to obtain L with the coloring μL . Set IL = IN . It is clear from the construction and from the lemmas that every prime interval of L is projective to a prime interval of IL and that distinct prime intervals of IL generate distinct join-irreducible congruences of L. It remains to see that if p and q are distinct prime intervals, then con(p) ≥ con(q) iﬀ pμL ≥ qμL . Since P is ﬁnite, it is suﬃcient to prove that con(p) con(q) in ConJ L iﬀ pμL qμL in J(D). But this is clear since if pμL qμL in J(D), then pμL = pi , for some 1 ≤ i ≤ m, and qμL = pij , for some 1 ≤ j ≤ ki , so con(p) con(q) was guaranteed in Ai . To establish that the size of L is O(n3 ), we give a very crude upper bound for |L|. Since 2n2 + 1 is an upper bound for |Ii |, 1 ≤ i ≤ m, it follows that 3(2n2 + 1) is an upper bound for |Ai | and 3(2n2 + 1)n is an upper bound

M

A N

Figure 9.8: Sketching the lattice L.

114

9. Semimodular Lattices

for |A|. Since |FA | ≤ n + 1, we get the upper bound (n + 1)2 + n + 1 for |M |. Finally, |IB | ≤ 2n2 + 1 + n + 1 = 2n2 + n + 2, so |N | ≤ 2(2n2 + n + 2)(n + 1). Therefore, 3(2n2 + 1)n + (n + 1)2 + n + 1 + 2(2n2 + n + 2)(n + 1) is an upper bound for L and it is a cubic polynomial in n. This completes the proof of the Theorem 9.1. It is not diﬃcult to ﬁnd better upper bounds for |L|; for instance, |L| ≤ 3n3 + 2n2 − 7n + 4.

9.4.

Discussion

An addendum In Chapter 13, we need a more detailed version of Theorem 9.1. Theorem 9.5. Let P be a ﬁnite order. Then there exists a ﬁnite semimodular lattice L with the following two properties: (i) ConJ L is isomorphic to P . (ii) |L| = O(|P |3 ). (iii) L has an congruence-determining sublattice C, which is an ideal and a chain. Problems In Chapter 8, the O(n2 ) construction is followed by a proof that this size is optimal. It would be nice to have a similar result for planar semimodular lattices. Problem 9.1. Is the O(n3 ) result optimal for planar semimodular lattices? Problem 9.2. Would the O(n3 ) result be optimal for semimodular lattices? Let Planar + SemiMod denote the class planar semimodular lattices. Using the notation of Section 7.5, these problems ask whether mcr(n, SemiMod) = O(n3 ) is optimal, and whether mcr(n, Planar + SemiMod) = O(n3 ) is optimal.

10 Modular Lattices

10.1.

The representation theorem

Corollary 3.9 notes that the congruence lattice of a ﬁnite modular lattice is boolean; however, a ﬁnite distributive lattice has a representation as the congruence lattice of an inﬁnite modular lattice by Schmidt [106] (see also Schmidt [112]): Theorem 10.1. Every ﬁnite distributive lattice D can be represented as the congruence lattice of a modular lattice L. We are going to prove this result in the following stronger form provided in Gr¨ atzer and Schmidt [74]. Theorem 10.2. Let P be a ﬁnite order. Then there exists a lattice L with the following properties: (i) L is a modular lattice. (ii) Con L is ﬁnite and ConJ L is isomorphic to P . (iii) L is countably inﬁnite. (iv) L is weakly atomic. (v) L has a unique complemented pair {a, a } = {0, 1} and id(a), id(a ) are chains, in fact, successor ordinals. (vi) L is rigid.

116

10. Modular Lattices

We are constructing a rigid lattice L, laying the foundation for the Independence Theorem for Modular Lattices; see Section 10.4. The lattice L we construct is inﬁnite, but “small.” As an order, it can be embedded in C 3 , where C is a countable ordinal; it is also weakly atomic.

10.2.

Proof-by-Picture

The gadgets in previous constructions were lattices with the three-element chain as the congruence lattice. The gadget in the modular case is the lattice M3 [C] of Chapter 5, for a chain C. We will use the sketch of this lattice as in Figure 5.1. Take the three-element order P = {p1 , p2 , p3 } with the only relation: p1 < p3 10.1. We present a Proof-by-Picture of Theorem 10.2 for the order P . We start by constructing the lattice L1 representing P1 = {p1 }, see Figure 10.2. The lattice L1 , clearly, is simple and has no automorphism. The congruence representing p1 is con(b11 , c11 ).

p3 p2

p1

Figure 10.1: The order P .

11 c11

q1

b11 a1

a1

01 Figure 10.2: The lattice L1 .

10.2. Proof-by-Picture

117

From L1 , we obtain the lattice L2 to represent {p1 , p2 }, the two-element antichain; see Figure 10.3. In L2 , we designated the elements b21 < c21 < b22 < c22 in the chain fil(a2 ), so that p1 is represented by con(b21 , c21 ) and p2 is represented by con(b22 , c22 ).

12 q2 c22 b22 c21 b21 a2

a2

11 q1

a1

a1

02

01

Figure 10.3: The lattice L2 .

Finally, we want to add p3 , so that it is over p1 but incomparable with p2 . This is accomplished in L3 ; see Figure 10.4, where L2 is only partially drawn. The lattice L3 is glued together from four parts; the boundaries are thick lines. The bottom part is L2 (only partially drawn and shaded), where on the upper left edge (the chain E = fil(a2 ) in L2 ) we mark the elements b21 < c21 <

118

10. Modular Lattices

13 c33

Top

b33

c32 b32 c31 b31

a3

a3

c22 b22

C

Left

C

Right

c21

D

b21 a2

a2

E

E

L2

Figure 10.4: The lattice L3 .

10.2. Proof-by-Picture

119

b22 < c22 , as described above. We take a chain C of type ω + 2, which we obtain by forming glued sums (introduced in Section 1.1.3) C = [b21 , c21 ] [b21 , c21 ] · · · ω-times and adding n0 ≺ n1 to the top. The left part of L3 , call it Left, and the right part of L3 , call it Right, are both C × E and the top part, call it Top, is M3 [C]. Finally, in Left, each square [b21 , c21 ] × [b21 , c21 ] is made into M3 [b21 , c21 ], and the top square of Left into a covering M3 . What are the congruences of L3 ? Think of L3 as the base B, a direct product of two chains, B = id(a3 ) × id(a3 ), with the ﬂaps of the M3 [X]-es sticking out (each X is a chain); then a congruence Θ of L3 is a congruence of B with the property that it acts the same way on the left edge (a chain) of an M3 [X] as on the right edge (an isomorphic chain). Therefore, every congruence of L2 extends, and extends uniquely to L3 . The extension of con(b22 , c22 ) (which corresponds to p2 ) is easy to see: in Left, for c, d ∈ C, c, x ≡ d, y iff c = d and x ≡ y in L2 . Similarly, in Right. No two distinct element are congruent in Top. The extension of con(b21 , c21 ) (which corresponds to p1 ) is much larger; since it collapses b21 and c21 , it collapses all of C; the quotient lattice is ﬁnite, as shown in Figure 10.5. Collapsing two distinct elements of {n0 , n1 } × E is equivalent to collapsing the two corresponding elements of E. If we collapse two elements of Left with the same C-component, this is obviously equivalent to collapsing the two Ecomponents. If we collapse two elements of Left with the same E-component, this is obviously equivalent to collapsing two elements of the left side of an M3 [b1 , c1 ], which makes it equivalent to the collapse of the corresponding two elements in the lower right edge, which is the last case. So there is exactly one new join-irreducible congruence, con(a2 , 0C , n1 , 0C ), which contains con(b21 , c21 ) but not con(b22 , c22 ). In L3 , we marked the elements b31 < c31 < b32 < c32 < b33 < c33 in the chain fil(a3 ), so that p1 is represented by con(b31 , c31 ), p2 is represented by con(b32 , c32 ), and p3 is represented by con(b33 , c33 ) in L3 . An automorphism α either ﬁxes a3 or takes it into a3 . But if a3 α = a3 , then α would deﬁne an automorphism of L2 that takes a2 into a2 , a contradiction. So α ﬁxes a2 and a2 . Therefore, α ﬁxes id(a2 ) and id(a2 ), and so id(a2 ) × id(a2 ). Since every element of L3 belongs to this direct product, or is on the third ﬂap of an M3 [X], whose base is in this direct product, we conclude that the base, and therefore, the whole M3 [X] is ﬁxed by α, so α is the identity map, that is, L3 is rigid.

120

10. Modular Lattices

Figure 10.5: The lattice L3 /con(b21 , c21 ).

10.3.

Construction and proof

We begin by constructing the basic building block of the construction, see Figure 10.6. Lemma 10.3. Let U be a chain with zero, 0U , and unit, 1U . Let 0U ≤ u < v ≤ 1U , and let V be the interval [u, v] of U . Construct the lattice L as follows: (i) Form the direct product U × V . (ii) Glue the interval [0U , u, u, v] of U × V with the ﬁlter [u, u, u, v] (∼ = V ) to M3 [V ] with the ideal { 0, 0, v | v ∈ V } ( ∼ = V ). (iii) Glue the lattice we obtained with the ﬁlter { 1, x, x | x ∈ V } ( ∼ = V) to the interval [v, u, 1U , v] with the ideal [v, u, v, v] ( ∼ = V ); the resulting lattice is denoted by U V .

10.3. Construction and proof

121

1U , v 1U , u

v, v M3 [V ] v, u U

u, v u, u

0U , v V 0U , u

Figure 10.6: The lattice U V . (iv) Consider U as a sublattice of U V by identifying x ∈ U with x, u if x ≤ u or if v ≤ x; otherwise, x ∈ [u, v], and we identify x with x, 0, 0 ∈ M3 [V ]. This identiﬁes U with a principal ideal of U V . Then every congruence Θ of U has a unique extension to a congruence Θ of U V ; therefore, Con U ∼ = Con(U V ). Proof. By Lemmas 2.13, 2.9, and 5.5. We use induction on n = |P |, the size of P , to construct a rigid modular lattice RP with the following properties: (I1 ) RP is a rigid modular lattice with zero, 0P , and unit, 1P . (I2 ) Every principal congruence of RP is a join of join-irreducible congruences and the join-irreducible congruences of RP form an order isomorphic to P . (I3 ) RP has an element aP with a unique complement aP such that the ideals id(aP ) and id(aP ) are isomorphic to a countable successor cardinal αP , and RP has a dual atom qP ≥ aP . (I4 ) Every congruence of RP is determined by its restriction to fil(aP ). Observe that (I4 ) is equivalent to the following condition: (I4 ) RP contains a chain P P P P P aP ≤ bP 1 < c1 ≤ b2 < c2 ≤ · · · ≤ bn < cn ≤ 1P

such that the join-irreducible congruences of RP are exactly the principal P P P P P congruences con(bP 1 , c1 ), con(b2 , c2 ), . . . , con(bn , cn ).

122

10. Modular Lattices

If |P | = 1, then the lattice of Figure 10.2 satisﬁes these requirements with αP = 4; this lattice is also simple and rigid. In general, if P is an antichain, we proceed as in the case n = 2, see Figure 10.3. We form the glued sum K = L1 L1 · · · L1 with n copies of L1 . Then we add elements (completing covering B2 -s to covering M3 -s) to the lattice C 2 , where C is a chain of length 3n, so that we can embed K into this lattice. It is clear that the lattice we obtain is a planar lattice satisfying (I1 )–(I4 ). Now let P be a ﬁnite order, not an antichain, and let us assume that RS has been constructed for all ﬁnite orders S satisfying |S| < |P |. Since P is not an antichain, we can choose a maximal element p of P that is not minimal. Let q1 , q2 , . . . , qr be the elements of P covered by p, with r ≥ 1. For the order Q = P − {p}, there exists a rigid modular lattice RQ such that (I1 )–(I4 ) hold; in particular, there is a chain Q Q Q Q Q aQ ≤ bQ 1 < c1 ≤ b2 < c2 ≤ · · · ≤ bn−1 < cn−1 ≤ 1Q ,

or simply, aQ ≤ b1 < c1 ≤ b2 < c2 ≤ · · · ≤ bn−1 < cn−1 ≤ 1Q , such that the join-irreducible congruences of RQ are the principal congruences con(b1 , c1 ), con(b2 , c2 ), . . . , con(bn−1 , cn−1 ). Let C = fil(a) in Q; by (I3 ) (observing that, by modularity, id(a ) ∼ = fil(a)), C is a well-ordered chain; it is isomorphic to αQ . Let con(bk1 , ck1 ), con(bk2 , ck2 ), . . . , con(bkr , ckr ) be the join-irreducible congruences corresponding to q1 , q2 , . . . , qr , respectively. Without loss of generality, we can assume that bk1 < ck1 ≤ bk2 < ck2 ≤ · · · ≤ bkr < ckr . Let F = [bk1 , ck1 ] [bk2 , ck2 ] · · · [bkr , ckr ]. For every natural number i, let Fi be a chain isomorphic to F , and let xi denote the image of x in Fi under this isomorphism. Finally, we consider the glued sum C of these chains, and the chain C that is C with two elements, n0 ≺ n1 , adjoined, C = F1 F2 · · · + {n0 , n1 } = C + {n0 , n1 }. We apply Lemma 10.3 ﬁrst to U1 = E and V1 = [bk1 , ck1 ], to obtain the lattice E [bk1 , ck1 ] and ﬁlter {ck1 } × E (∼ = E), and second to U2 = E and V2 = [bk2 , ck2 ], in order to obtain the lattice E [bk2 , ck2 ] and ideal {bk2 } × E (∼ = E); and we glue these two lattices together over the given ideal and ﬁlter.

10.3. Construction and proof

123

We proceed similarly and glue E [bk3 , ck3 ] to the resulting lattice, and so on. In r steps, we obtain a lattice L with an ideal and a ﬁlter both isomorphic to E. Now for each i < ω, we take a copy Li of L, and glue L2 to L1 , L3 to the resulting lattice, and so on; in the last step, we glue Li . Call the lattice we obtained Li . Obviously, L1 ⊆ L2 ⊆ · · · ⊆ Ln , so we can take ( Li | i < ω ). We adjoin to this lattice {n0 , n1 } × E (where {n0 , n1 } is regarded as the two-element chain with n0 < n1 ) so that u, c ∧ v, d = u ∧ v, c ∧ d, u, c ∨ v, d = u ∨ v, c ∨ d, for u and/or v ∈ {n0 , n1 } and c, d ∈ E. We add one more element. E has a dual atom qC by (I3 ); the lattice {n0 , n1 }×{qC , 1Q } is a cover-preserving four-element boolean sublattice of the lattice we have constructed. We add an element w so that {n0 , n1 } × {qC , 1Q } with w form a cover-preserving M3 . Let Left denote the lattice we have just obtained. Finally, let us glue RQ and Left together over the ﬁlter E and ideal {bk1 }× E (∼ = E) in L0 ; call the resulting lattice Left+ . To investigate the lattice Left, let us present a more intuitive description. Form the direct product of C and E; we will call this the base of Left. Consider the interval [(bkj )i , (ckj )i ] in C and the corresponding interval [bkj , ckj ] in E. In the direct product C × E, replace [bikj , cikj ] × [bkj , ckj ] with M3 [bkj , ckj ] so that x, y ∈ [bikj , cikj ] × [bkj , ckj ] is replaced by x, x ∧ y, y and we add the element w in the top prime square so that it form an M3 . Of course, with this deﬁnition it is not clear whether Left is a lattice, whether it is modular, and what are the congruences of Left. It is clear, however, that this deﬁnition is the same as the more complicated one given above, so we get all these properties of the construction from the results in Chapter 5 and from Lemma 10.3. Take a congruence Ψ of Left; then the restriction of Ψ to the base of Left is of the form Φ × Θ, where Φ is a congruence of C and Θ is a congruence of E. It is clear from Lemma 10.3 that Φ and Θ uniquely determine Ψ. Moreover, Θ “almost” determines Φ. In fact, a ≡ b (Φ) is fully determined by Θ, for a, b ∈ C. The same is true for a, b ∈ {n0 , n1 }. Finally, let a ∈ C and b ∈ {n0 , n1 }. Then a ≡ b (Φ) implies that all Di are collapsed by Φ from some i on, and so all of C is collapsed by Φ. We conclude that Left has exactly one join-irreducible congruence that is not a minimal extension

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of a congruence from E, namely, con(0C , 0E , n0 , 0E ), and this congruence majorizes all the join-irreducible congruences that are minimal extensions of the congruences con(b1 , c1 ), con(b2 , c2 ),. . . , con(bn−1 , cn−1 ) of E to Left. By the inductive assumption (I3 ), every congruence of RQ is determined by its action on E, and so we obtain that every congruence of RQ can be extended to Left+ . Therefore, the join-irreducible congruences of Left+ can be described as follows: they are the minimal extensions of the join-irreducible congruences of RQ to Left+ and the congruence con(0C , 0E , n0 , 0E ). Hence they form an order isomorphic to P , and so ConJ Left+ ∼ = P , by (I2 ). The lattice Left+ does not satisfy all the inductive assumptions, so we will deﬁne RP as an appropriate extension of Left+ . Deﬁne the lattice Top as M3 [C] and deﬁne the ideal C1 = { 0C , 0C , x | x ∈ C } of M3 [C]. Deﬁne the lattice Right = [aQ , 1Q ] × C1 ∼ C) be the ﬁlter generated by 1Q , 0C in Right. Let Right+ be and let C2 (= 1 the gluing of Top and Right over C1 and C2 . Obviously, the ideal of Right+ generated by 1C , 0C , 0C ∈ Top is isomorphic with the ﬁlter fil(aQ ) of Left+ . So we can glue Left+ and Right+ together to obtain RP . Deﬁne aP = n1 , 0E (∈ Left) and let aP be its unique complement, aP = aQ , 0C1 (∈ Right). The dual atom n1 , n0 , n0 of Top can serve as the dual atom of RP . Now we verify that RP satisﬁes the conditions of Theorem 10.2 as well as the inductive assumptions (I1 )–(I4 ). id(aP ) is a countable well-ordered chain, namely, αQ C, which is a successor ordinal; the same is true of id(aP ). Let α be an automorphism of RP . Since 1Q is the smallest element x ∈ RP such that there is a sublattice isomorphic to M3 from x to 1P , it follows that 1Q α = 1Q . Since RQ is rigid, it follows that xα = x, for all x ∈ RQ ⊆ RP . Left is built on C × E, and E is kept ﬁxed by α. Therefore, α maps C into itself. But C is well-ordered, so α is the identity map on C, and so on all of Left. Top = [1Q , 1P ] and both these elements are ﬁxed by α. Next, consider the three atoms in the sublattice isomorphic to M3 , stretching from 1Q to 1P : the element 1C , 0C , 0C is ﬁxed because it is in RQ ; the elements 0C , 1C , 0C and 0C , 0C , 1C cannot be interchanged because 1Q has a complement in the interval [aQ , 0C , 0C , 1C ] but not in [aQ , 0C , 1C , 0C ]. So α ﬁxes the elements 1C , 0C , 0C , 0C , 1C , 0C , and 0C , 0C , 1C and also the chain C, hence α ﬁxes all of the lattice Top. Since we already know that in Right, α ﬁxes id(1Q ) and fil(1Q ), it follows that α ﬁxes every element of Right. We conclude that RP is rigid.

10.4. Discussion

125

It remains to verify (I4 ). In RQ , we were given the chain aQ ≤ b1 < c1 ≤ b2 < c2 ≤ · · · ≤ bn < cn ≤ 1Q such that the join-irreducible congruences of RQ are the principal congruences con(b1 , c1 ), con(b2 , c2 ), . . . , con(bn−1 , cn−1 ). In RP , we deﬁne bP i = aQ ∨ bi , cP i = aQ ∨ ci , for i = 1, . . . , n − 1, and bP n = 1C , 0C , 0C , ∈ Top, cP n = 1P ∈ Top. Then condition (I4 ) is obvious for the chain P P P P P aP ≤ bP 1 < c1 ≤ b2 < c2 ≤ · · · ≤ bn < cn = 1P .

This completes the proof of Theorem 10.2.

10.4.

Discussion

The Independence Theorem for Modular Lattices In [24] (Problem II. 18), I raised the problem whether the congruence lattice and the automorphism group of a ﬁnite lattice are independent. This problem was solved aﬃrmatively by Baranski˘ı [3], [4] and Urquhart [120]. This topic is discussed in Chapter 15 of this book. It is natural to ask whether one could prove the Independence Theorem for modular lattices. This was done in Gr¨ atzer and Schmidt [74]. Theorem 10.4 (Independence Theorem for Modular Lattices). Let D be a nontrivial ﬁnite distributive lattice and let G be a ﬁnite group. Then there exists a modular lattice M such that the congruence lattice of M is isomorphic to D and the automorphism group of M is isomorphic to G. The proof uses the following statement: Theorem 10.5. Let G be a ﬁnite group. Then there exists a simple modular lattice S with an atom p such that the automorphism group of S is isomorphic to G and every automorphism keeps p ﬁxed. A weaker form of this theorem (just constructing a modular lattice) is due to Mendelsohn [98]. The present form is in Gr¨ atzer and Schmidt [70]; see also Herrmann [88] and Schmidt [113].

126

10. Modular Lattices

Now it is easy to prove the Independence Theorem for Modular Lattices. Let G and D be given as in the Independence Theorem. Let S be constructed as in Theorem 10.5 for G and let R be the lattice in Theorem 10.2 constructed for D. We take the ideal id(p) in S and the ﬁlter fil(q) in R; both are twoelement lattices, so we can glue S and R together over id(p) and fil(q). Let M be the resulting lattice. M is a modular lattice by Lemma 2.24. By Lemma 2.9, Con M ∼ = Con R, so Con M is isomorphic to D. Now let α be an automorphism of S, and deﬁne a map α† of M into itself as follows: xα, if x ∈ S; † xα = x, if x ∈ R. Since p and 0S are kept ﬁxed by α, it is evident that α† is an automorphism of M . Moreover, q and p is the only pair of elements in M satisfying q ≺ p and M = id(p) ∪ fil(q); therefore, every automorphism of M acts as an automorphism on S and R. It is now clear that every automorphism of M is of the form α† for some automorphism α of S, hence, α → α† is an isomorphism between the automorphism group of S and the automorphism group of M . It follows that the automorphism group of M is isomorphic to G. This completes the proof of the Independence Theorem for Modular Lattices. Two stronger results Two results state Theorem 10.1 in stronger form. Theorem 10.6 (Freese [15]). Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnitely generated modular lattice L. Theorem 10.7 (Schmidt [110]). Every ﬁnite distributive lattice D can be represented as the congruence lattice of a complemented modular lattice L. Arguesian lattices Actually, in Gr¨ atzer and Schmidt [74], we prove a much stronger form of the Independence Theorem: We construct an arguesian lattice A with a given ﬁnite congruence lattice and ﬁnite automorphism group. Recall that a lattice is arguesian if it satisﬁes the arguesian identity: (x1 ∨ y1 ) ∧ (x2 ∨ y2 ) ∧ (x3 ∨ y3 ) ≤ ((z ∨ x2 ) ∧ x1 ) ∨ ((z ∨ y2 ) ∧ y1 ). This is a much stronger form of modularity; it is a lattice theoretic form of Desargues’ Theorem. I would recommend that the reader consult Sections IV.4 and IV.5 of [26] for background.

10.4. Discussion

127

To prove the Independence Theorem for Arguesian Lattices, we need arguesian versions of Theorems 10.2 and 10.5. In fact, in Gr¨ atzer and Schmidt [70], we accomplished that for Theorem 10.5. In Gr¨ atzer and Schmidt [74], we prove that the lattice constructed in Section 10.3 is arguesian, and that the two arguesian theorems can be combined to get the Arguesian Independence Theorem. Problems Problem 10.1. Is there an analogue of Theorem 10.1 for n-distributive modular lattices. Part IV of this book deals with congruence-preserving extensions. Modular lattices are conspicuous with their absence in that part. Here are some obvious problems to raise: Problem 10.2. Does every modular lattice have a proper, modular, congruence-preserving extension? Problem 10.3. Let L be a modular lattice with a ﬁnite congruence lattice and let G be a group. Does L have a modular congruence-preserving extension whose automorphism group is isomorphic to G? In light of Theorem 10.7, we can ask: Problem 10.4. Does the Independence Theorem for Complemented Modular Lattices hold? Problem 10.5. Are there any results (paralleling the results of Chapters 11 and 14) about regular, uniform, or isoform modular lattices? Nothing is really known about congruence lattices of inﬁnite modular lattices. It is astonishing that the following problem seems open: Problem 10.6. Is it true that for every lattice L, there is a modular lattice M with Con L ∼ = Con M ? Can Freese’s Theorem 10.6 and Schmidt’s Theorem 10.7 be combined? Problem 10.7. Can every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnitely generated, complemented, modular lattice L?

11 Uniform Lattices

11.1.

The representation theorem

Why couldn’t lattices be more like groups and rings? We want representation theorems by lattices whose congruences behave “as in groups and rings.” We will look at lattices that are regular and uniform. Let L be a lattice. We call a congruence relation Θ of L regular if any congruence class of Θ determines the congruence. Let us call the lattice L regular if all congruences of L are regular. Sectionally complemented lattices are regular, so Theorem 7.5 is a representation theorem for ﬁnite regular lattices. In this chapter we consider a concept stronger than regularity. We call a congruence relation Θ of a lattice L uniform if any two congruence classes of Θ are of the same size (cardinality). Let us call the lattice L uniform if all congruences of L are uniform. The following result was proved in Gr¨ atzer, Schmidt, and Thomsen [79]: Theorem 11.1. Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnite uniform lattice L. A uniform lattice is always regular, so the lattices of Theorem 11.1 are also regular. Figure 11.1 shows the result of the construction for D = C4 .

11.2.

Proof-by-Picture

Let A be a ﬁnite lattice. We introduce a new lattice, A+ .

130

11. Uniform Lattices

Figure 11.1: The uniform construction for the four-element chain. Form A × C2 . Then for every a ∈ A, we get a prime interval [a, 0, a, 1]. We replace this by a copy of B2 , for all a ∈ A, to obtain A+ . We identify a ∈ A with a, 0 and b ∈ B2 with the copy of b in the interval [0, 0, 0, 1]; so we can regard A and B2 as sublattices of A+ . Figure 11.2 shows the diagrams of B1+ and B2+ . It is easy to compute that A+ is a subdirectly irreducible lattice with base congruence con(0, 0, 0, 1), and A+ /con(0, 0, 0, 1) is isomorphic to A; the isomorphism provided by a → a, 0/con(0, 0, 0, 1). So Con A+ is isomorphic to Con A with a new zero added. We present the Proof-by-Picture of Theorem 11.1 in the following form. Theorem 11.2. For any ﬁnite order P , there exists a ﬁnite uniform lattice L such that ConJ L is isomorphic to P , and L satisﬁes the following property: (A) If Θ1 , Θ2 , . . . , Θn ∈ ConJ L are pairwise incomparable, then L contains atoms p1 , p2 , . . . , pn that generate an ideal isomorphic to Bn and satisfy

11.2. Proof-by-Picture

B1

B2

B2

131

B2

Figure 11.2: The lattices B1+ and B2+ . Θi = con(0, pi ), for all i ≤ n. Let P be a ﬁnite order with n elements. If n = 1, then D ∼ = B1+ ; see Figure 11.2. Let us assume that, for all ﬁnite distributive lattices with fewer than n join-irreducible elements, there exists a lattice L satisfying Theorem 11.2. Let q be a minimal element of P and let q1 , . . . , qk (k ≥ 0) list all upper covers of q in P . Let P1 = P − {q}. By the inductive assumption, there exists a lattice L1 satisfying ConJ L1 ∼ = P1 and condition (A). If k = 0, then we deﬁne L = B1+ × L1 . So we assume that 1 ≤ k. The congruences of L1 corresponding to the qi ’s are pairwise incomparable and therefore can be written in the form con(0, pi ) so that the pi ’s generate an ideal I isomorphic to Bk . Figure 11.3 shows the inductive step. The lattices L1 and I + share the ideal I, so we can form the chopped lattice K = Merge(L1 , I + ), in the notation of Section 4.1. We deﬁne L = Id K.

L1

I+ I

Figure 11.3: The inductive step: the chopped lattice K.

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11. Uniform Lattices

The idea is that con(0, 0, 0, 1) is a join-irreducible congruence of L, below the con(0, pi ), for all i ≤ n, but not below any other minimal join-irreducible congruence of L1 . And there is no other new join-irreducible congruence in L, so ConJ L ∼ = P. Of course, one has to compute that L is uniform, which is tedious. The induction condition (A) is easy to verify for L.

11.3.

The lattice N (A, B)

We introduce the A+ construction of Section 11.2 in a more general form; see Section 11.5 for an even more general lattice construction. The construction Let A and B be lattices. Let us assume that A is bounded, with bounds 0 and 1, with 0 = 1. Let A− = A − {0, 1}. We introduce a new lattice construction N (A, B)—the A+ of Section 11.2 is the special case N (A, B2 ). For u ∈ A × B, we use the notation u = uA , uB ; the binary relation ≤× will denote the ordering on A × B, and ∧× , ∨× the meet and join in A × B, respectively. On the set A × B, we deﬁne a binary relation ≤N (A,B) (denoted by ≤N if A and B are understood) as follows: ≤N = ≤× − { u, v | u, v ∈ A− × B, uB = vB }. We deﬁne N (A, B) = A × B, ≤N . Lemma 11.3. The set N (A, B) is ordered by ≤N , in fact, N (A, B) is a lattice. The join and meet in N (A, B) of ≤N -incomparable elements can be computed using the formulas: 1, uB ∨ vB , if u ∨× v ∈ A− × B and uB = vB ; u ∨N v = otherwise. u ∨× v, 0, uB ∧ vB , if u ∧× v ∈ A− × B and uB = vB ; u ∧N v = u ∧× v, otherwise. Proof. Since ≤× is reﬂexive, it follows that ≤N is reﬂexive, since uB = uB fails, for all u ∈ A × B. Since ≤× is antisymmetric, so is ≤N . Let u, v, w ∈ A × B; let us assume that u ≤N v and v ≤N w. Since ≤× is transitive, we conclude that u ≤× w. So if u ≤N w fails, then u, w ∈ A− × B and uB = wB . It follows that v ∈ A− × B and either uB = vB or vB = wB , contradicting that u ≤N v or v ≤N w. So ≤N is transitive, proving that N (A, B) is an order. We prove that N (A, B) is a lattice by verifying the join and meet formulas. By duality, we only have to verify one of them; we will prove the meet formula.

11.3. The lattice N (A, B)

133

Let u, v ∈ A×B be ≤N -incomparable elements, and let t be a lower bound of u and v in N (A, B). Case 1. u ∧× v is not a lower bound of both u and v in N (A, B). If u∧× v is not a lower bound of both u and v in N (A, B), say, u∧× v N u, then u, u ∧× v ∈ A− × B and uB = (u ∧× v)B (which is the same as uB = vB ). Since t ≤× u ∧× v, it follows that tB ≤ (u ∧× v)B < uB , so t ∈ / A− × B (otherwise, we would have t N u). We conclude that t = 0, tB , so tB ≤ uB ∧ vB , which yields that t ≤ 0, uB ∧ vB . So in Case 1, u ∧N v = 0, uB ∧ vB . Case 2. u ∧× v is a lower bound of both u and v in N (A, B). If t N u∧× v, then t, u∧× v ∈ A− ×B and tB < (u∧× v)B , so u ∈ A− ×B or v ∈ A− ×B, say, u ∈ A− ×B. Therefore, the assumption of Case 2, namely, u ∧× v ≤N u, implies that (u ∧× v)B = uB . So t, u ∈ A− × B and tB = uB , contradicting that t ≤N u. Thus t N u ∧× v leads to a contradiction. We conclude that t ≤N u ∧× v. So in Case 2, u ∧N v = u ∧× v. Since the two cases correspond to the two clauses of the meet formula, this veriﬁes the meet formula. We will use the notation: B∗ = {0} × B, B ∗ = {1} × B, and for b ∈ B, Ab = A × {b}. Note that B∗ is an ideal and B ∗ is a ﬁlter of N (A, B). Congruences Let K and L be lattices, and let α be an embedding of K into L. Given a congruence Θ of L, we can deﬁne the congruence Θ1 on K via α, that is, for a, b ∈ K, a ≡ b (Θ1 ) iff aα ≡ bα (Θ). We will call Θ1 the restriction of Θ transferred via the isomorphism α to K. Let Ψ be a congruence relation of N = N (A, B). Using the natural isomorphisms of B into N (A, B) with images B∗ and B ∗ , we deﬁne Φ∗ and Φ∗ as the restriction of Ψ to B∗ and B ∗ , respectively, transferred via the natural isomorphisms to B. Let Θb as the restriction of Ψ to Ab , for b ∈ B, transferred via the natural isomorphisms to A. Lemma 11.4. Φ∗ = Φ∗ . Proof. Indeed, if b1 ≡ b2 (Φ∗ ), then 0, b1 ≡ 0, b2 (Ψ). Joining both sides with 1, b1 ∧ b2 , we obtain that 1, b1 ≡ 1, b2 (Ψ), that is, b1 ≡ b2 (Φ∗ ). This proves that Φ∗ ≤ Φ∗ . We prove the reverse inclusion symmetrically. It is easy to see that Φ = Φ∗ = Φ∗ ∈ Con B, and { Θb | b ∈ B } ⊆ Con A describe Ψ, but it is diﬃcult to obtain a description of the congruences of N (A, B), in general. So we impose a very stringent condition on A.

134

11. Uniform Lattices

Lemma 11.5. Let A and B be lattices with |A| > 2 and |B| > 1; let A be bounded, with bounds 0 and 1. Let us further assume that A is non-separating. Then the map sending Ψ = ωN to its restriction to B∗ transferred to B by the natural isomorphism is a bijection between the non-ωN congruences of N (A, B) and the congruences of B. Therefore, Con N (A, B) is isomorphic to Con B with a new zero added. Proof. Let Ψ = ωN be a congruence relation of N (A, B). We start with the following statement: Claim 11.6. There are elements a1 < a2 in A and an element b1 ∈ B such that a1 , b1 ≡ a2 , b1 (Ψ). Proof. Assume that u1 , v1 ≡ u2 , v2 (Ψ) with u1 , v1 2, we can pick an a ∈ A− . Then a, v1 = a, v1 ∨ 0, v1 ≡ a, v1 ∨ 0, v2 = 1, v2 (Ψ), from which we conclude that a, v1 ≡ 1, v1 (Ψ), so the claim is veriﬁed with a1 = a, a2 = 1, and b1 = v1 . Second case. u1 < u2 . Since we have assumed that u1 , v1
11.3. The lattice N (A, B)

135

Since A is non-separating, there exists a3 ∈ A with 0 < a3 and 0 ≡ a3 (con(a1 , a2 )). Moreover, Ab1 is a sublattice of N (A, B), so it follows that 0, b1 ≡ a3 , b1 (con(a1 , b1 , a2 , b1 )), and thus 0, b1 ≡ a3 , b1 (Ψ). Therefore, for any b2 ∈ B with b1 < b2 , joining both sides with 0, b2 , we obtain that 0, b2 ≡ 1, b2 (Ψ), that is, Ab2 is in a single Ψ class. This completes the proof of the claim, unless b1 is the unit element, 1B , of B. In this case, 0, 1B ≡ a3 , 1B (Ψ). Since A is non-separating, there exists a4 ∈ A with a4 < 1 and a4 ≡ 1 (con(0, a3 )). Moreover, A1B is a sublattice of N (A, B), so it follows that a4 , 1B ≡ 1, 1B (con(0, 1B , a3 , 1B )), and thus a4 , 1B ≡ 1, 1B (Ψ). Now choose any b2 < 1B (recall that we have assumed that |B| > 1). Meeting both sides with 1, b2 , we obtain that 1, b2 ≡ 0, b2 (Ψ), that is, Ab2 is in a single congruence class of Ψ. Claim 11.8. Ab is in a single congruence class of Ψ, for each b ∈ B. Proof. Let b ∈ B. By Claim 11.7, there is an element b2 ∈ B such that Ab2 is in a single congruence class of Ψ, that is, 1, b2 ≡ 0, b2 (Ψ). Therefore, 1, b = (1, b2 ∨ 0, b ∨ b2 ) ∧ 1, b ≡ (0, b2 ∨ 0, b ∨ b2 ) ∧ 1, b = 0, b (Ψ), that is, Ab is in a single congruence class of Ψ. Now the statement of the lemma is easy to verify. It is clear that the map Ψ → Φ is one-to-one, for Ψ ∈ Con N (A, B) − {ωN }. It is also onto: given a congruence Φ of B, deﬁne Ψ on N (A, B) by u1 , v1 ≡ u2 , v2 (Ψ)

iff v1 ≡ v2 (Φ).

Then Ψ is in Con N (A, B) − {ωN } and it maps to Φ.

136

11. Uniform Lattices

Congruence classes Now let U be a ﬁnite lattice with an ideal V isomorphic to Bn . We identify V with the ideal (Bn )∗ = id(0, 1) of N (B2 , Bn ) to obtain the merged chopped lattice K = Merge(U, N (B2 , Bn )) (using the notation of Section 4.1). Let m denote the generator of V = (Bn )∗ . We view Id K as the set of compatible pairs u, y with u ∈ U and y ∈ N (B2 , Bn ) (see Section 4.2). Lemma 11.9. Let u ∈ U . Then { y ∈ N (B2 , Bn ) | u, y ∈ Id K } is isomorphic to B2 . Proof. This is clear since there are exactly four elements y of N (B2 , Bn ) satisfying that u ∧ m = y ∧ m, namely, the elements of (B2 )u∧m , which form a sublattice isomorphic to B2 . Therefore, { y ∈ N (B2 , Bn ) | u, y ∈ Id K } is a four-element set, closed under coordinate-wise meets and joins; the statement follows. Now let us further assume that U is uniform. Lemma 11.10. Id K is uniform. Proof. A congruence Λ of Id K can be described (see Section 4.3) as a congruence vector Θ, Ψ, where Θ is a congruence of U , Ψ is a congruence of N (B2 , Bn ), and Θ and Ψ restrict to the same congruence of V = (Bn )∗ . The trivial congruences, ωId K (represented by ωU , ωN (B2 ,Bn ) ) and ιId K (represented by ιU , ιN (B2 ,Bn ) ), are obviously uniform. Therefore, we need only look at two cases. First case. Λ is represented by Θ, ω. So ΘV = ωV . Let x, y be an element of Id K and note that x, y/Θ, ω = { t, y ∈ Id K | t ≡ x (Θ) }. If t ≡ x (Θ), then t ∧ m ≡ x ∧ m (Θ), but ΘV = ωV , so t ∧ m = x ∧ m. We conclude that x, y/Θ, ω = { t, y | t ≡ x (Θ) }, and |x, y/Θ, ω| = |x/Θ|.

11.4. Formal proof

137

So Λ is uniform; each congruence class of Λ is of the same size as a congruence class of Θ. Second case. Λ is represented by Θ, Ψ, where Ψ = ω. Let x, y be an element of Id K. Then x, y/Θ, Ψ = { w, z ∈ Id K | x ≡ w (Θ) and y ≡ z (Ψ) }. For a given w if w, t1 and w, t2 ∈ Id K, then t1 ≡ t2 (Ψ) because (B2 )w is in a single congruence class of Ψ by Lemma 11.5 (in particular, by Claim 11.8) and { t ∈ N (B2 , Bn ) | w, t ∈ Id K } = (B2 )w∧m ; thus |{ t ∈ N (B2 , Bn ) | w, t ∈ Id K }| = |(B2 )w∧m | = 4. We conclude that x, y/Θ, Ψ = { w, z ∈ Id K | x ≡ w (Θ), z ∈ (B2 )w∧m }, and |x, y/Θ, Ψ| = 4|x/Θ|. So Λ is uniform; each congruence class of Λ is four-times the size of a congruence class of Θ.

11.4.

Formal proof

Based on the results of Section 11.3, we will prove Theorem 11.2 by induction. Let P be a ﬁnite order with n elements. If n = 1, then Down P ∼ = B1 , so there is a lattice L = B1 that satisﬁes Theorem 11.2. Let us assume that for all ﬁnite orders with fewer than n elements, there exists a lattice L satisfying Theorem 11.2. Let q be a minimal element of P and let q1 , . . . , qk (k ≥ 0) list all upper covers of q in P . Let P1 = P − {q}. By the inductive hypothesis, there exists a lattice L1 satisfying ConJ L1 ∼ = P1 and also satisfying condition (A). If k = 0, then Down P ∼ = B1 × Down P1 , and so L = B1 × L1 obviously satisﬁes all the requirements of the theorem. So we assume that 1 ≤ k. The congruences of L1 corresponding to the qi ’s are pairwise incomparable; therefore, these congruences can be written in the form con(0, pi ) and the pi ’s generate an ideal I1 isomorphic to Bk . The lattice N (B2 , Bk ) also contains an ideal (Bk )∗ isomorphic to Bk . Merging these two lattices by identifying I1 and (Bk )∗ , we get the chopped lattice K and the lattice L = Id K. The chopped lattice K is sketched in Figure 11.4. L is uniform by Lemma 11.10.

138

11. Uniform Lattices

L1

I1 = (Bk )∗

N (B2 , Bk )

Figure 11.4: The chopped lattice K for the formal proof.

Let Θ be a join-irreducible congruence of L. Then we can write Θ as con(a, b), where a is covered by b. We can assume that either a, b ∈ L1 or a, b ∈ N (B2 , Bk ). In either case, we ﬁnd an atom q in L1 or in N (B2 , Bk ), so that con(a, b) = con((0, q) in L1 or in N (B2 , Bk ). Obviously, q is an atom of L and con(a, b) = con(0, q) in L. Let Θ1 , Θ2 , . . . , Θt be pairwise incomparable join-irreducible congruences of L. To verify condition (A), we have to ﬁnd atoms p1 , p2 , . . . , pt of L satisfying Θi = con(0, pi ), for all i ≤ t, and such that p1 , p2 , . . . , pt generate an ideal of L isomorphic to Bt . Let p denote an atom in N (B2 , Bk ) − I1 ; there are two, but they generate the same congruence con(0, p). If con(0, p) is not one of Θ1 , Θ2 , . . . , Θt , then clearly we can ﬁnd p1 , p2 , . . . , pt in L1 as required, and p1 , p2 , . . . , pt also serves in L. If con(0, p) is one of Θ1 , Θ2 , . . . , Θt , for instance, con(0, p) = Θt , then let p1 , p2 , . . . , pt−1 be the set of atoms establishing condition (A) for the congruences Θ1 , Θ2 , . . . , Θt−1 in L1 and therefore, in L. Then the elements p1 , p2 , . . . , pt−1 , p represent the congruences Θ1 , Θ2 , . . . , Θt and generate an ideal isomorphic to Bt . Therefore, L satisﬁes condition (A). Finally, it is clear from this discussion that ConJ K has exactly one more element than ConJ L1 , namely, con(0, p), and this congruence relates to the congruences in ConJ L1 exactly as q relates to the elements of P . Therefore, P ∼ = ConJ L.

11.5. Discussion

11.5.

139

Discussion

Isoform lattices The next logical step is to require that that any two congruence classes be not only of the same size but isomorphic as lattices. We called such lattices isoform. The following result was proved in Gr¨ atzer and Schmidt [76]: Theorem 11.11. Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnite isoform lattice L. Since isomorphic lattices are of the same size, Theorem 11.11 is a much stronger version of Theorem 11.1. Figure 11.1 shows that the lattice we obtained in this chapter for D = C4 is not isoform. Indeed, the congruence marked by the two dotted ovals has two congruence classes of 16 elements each, but they are not isomorphic; in the top lattice, two dual atoms are join-irreducible, and in the bottom lattice, none are join-irreducible. Note that isoform lattices are not “like groups and rings.” The isoform concept is special to idempotent algebras. The N (A, B, Θ) construction We could prove Theorem 11.11 based on the following generalization of the N (A, B) construction. Let P = P, ≤P be a ﬁnite order. Then the ordering ≤P on P is the reﬂexive-transitive closure of ≺P , the covering relation in P, ≤P , in formula: rt(≺P ) = ≤P . Now take a subset H of ≺P , and take the reﬂexive-transitive extension rt(H) of H. Then P, rt(H) is also an order; we call it a pruning of P . If you think of P in terms of its diagram, then the terminology is easy to picture: We obtain the diagram of P, rt(H) from the diagram of P by cutting out (pruning) some edges (each representing a covering) but not deleting any elements. For instance, the lattice of Figure 11.1 is a pruning of the boolean lattice B25 . Let L be a ﬁnite lattice. We call L discrete-transitive if for any congruence Φ of L and for a < b < c in L, whenever Φ is discrete on [a, b] and [b, c], then Φ is discrete on [a, c]. Let A be a nontrivial ﬁnite lattice with bounds 0 and 1; let |A| > 2. Set A− = A − {0, 1}. Let B be a nontrivial ﬁnite lattice with a discrete-transitive congruence Θ. To prune A × B, we deﬁne the set: Prune(A, B, Θ) = { a, b1 , a, b2 | a ∈ A− , b1 ≺ b2 in B, b1 ≡ b2 (Θ) }. Prune(A, B, Θ) is a subset of ≺× , so we can deﬁne H = ≺× − Prune(A, B, Θ). Now we take the reﬂexive-transitive extension rt(H) of H. The set A × B with the ordering rt(H) is N (A, B, Θ). It is clear that N (A, B, ω) is the direct product A × B and N (A, B, ι) = N (A, B).

140

11. Uniform Lattices

We describe in Gr¨ atzer and Schmidt [76] the order N (A, B, Θ): u ≤ v in N (A, B, Θ) iff uA , vA ∈ A− and [uB , vB ] is Θ-discrete, or

uA or vA ∈ / A− ,

and we prove that N (A, B, Θ) is a lattice under this ordering. Pruning seldom produces a lattice. Note the very strong condition imposed on Θ to make the pruned order a lattice. We use the N (A, B, Θ) construction to obtain the following version of Theorem 11.11 (Gr¨ atzer and Schmidt [76]): Theorem 11.12. Every ﬁnite distributive lattice D can be represented as the congruence lattice of a ﬁnite lattice L with the following properties: (i) L is isoform. (ii) For every congruence Θ of L, the congruence classes of Θ are projective intervals. (iii) L is a ﬁnite pruned boolean lattice. (iv) L is discrete-transitive. By Properties (i) and (ii), for every congruence relation Θ of L and for any two congruence classes U and V of Θ, the congruence classes U and V are required to be isomorphic and projective intervals, but we do not require that there be a projectivity that is also an isomorphism. We refer the reader to Gr¨ atzer and Schmidt [76] for a proof of Theorem 11.12. A stronger form of Theorem 11.12 will be proved in Chapter 14. Problems Let us start with some general problems on uniformity. Problem 11.1. Develop a theory of uniform lattices. Problem 11.2. Prove that uniformity is not a ﬁrst-order property. The answer to the following question is likely to be in the negative. Problem 11.3. Are inﬁnite relatively complemented lattices uniform? The congruence classes of a congruence of the lattice L in Theorem 11.12 are both isomorphic and projective. Problem 11.4. Is it possible to sharpen Theorem 11.12 so that the isomorphisms of the congruence classes be projectivities?

11.5. Discussion

141

Can we combine the main result of this chapter with the results of the previous chapters? In other words, Problem 11.5. Can every ﬁnite distributive lattice D be represented as the congruence lattice of a ﬁnite uniform (resp., isoform) lattice L with some additional property: L be semimodular, sectionally complemented, or 2-distributive, and so on? Problem 11.6. Can every ﬁnite distributive lattice D be represented as the congruence lattice of a uniform (resp., isoform) modular lattice L? What happens in the inﬁnite case: Problem 11.7. Is there an analogue of Theorem 11.1 (resp., Theorem 11.11) for inﬁnite lattices? Let Uniform and Isoform denote the class of uniform and isoform lattices, respectively. (Recall that the function mcr was introduced in Section 7.5.) Problem 11.8. What is mcr(n, Uniform)? What is mcr(n, Isoform)? Can one get a good result for O(mcr(n, Uniform)) and O(mcr(n, Isoform))? Let L be a ﬁnite uniform lattice and let ϕ be an isomorphism between a ﬁnite distributive lattice D and Con L. Then we can introduce a function s = s(L, ϕ) from D to the natural numbers, as follows: Let d ∈ D; then dϕ is a congruence of L. Since L is uniform, all congruence classes of dϕ are of the same size. Let s(d) be the size of the congruence classes. The function s has the following obvious properties: (s1 ) s(0) = 1. (s2 ) If a < b in D, then s(a) < s(b). Problem 11.9. Characterize the function s. In other words, let f be a function from a ﬁnite distributive lattice D to the natural numbers that satisﬁes conditions (s1 ) and (s2 ) above. What additional conditions on the function f are required for f = s(L, ϕ), for some ﬁnite uniform lattice L and isomorphism ϕ : D → Con L? This problem may be too diﬃcult to solve in its full generality. The following lists some special cases that may be easier to attack. Problem 11.10. Characterize the function s for some special classes of ﬁnite distributive lattices: (i) ﬁnite boolean lattices; (ii) ﬁnite chains; (iii) “small” distributive lattices.

Part IV

Extensions

12 Sectionally Complemented Lattices

12.1.

The extension theorem

As outlined in the preface, Part IV deals with congruence-preserving extension theorems. We start with sectionally complemented lattices, as we began Part III. The reader should compare the relatively small constructs and shorter proofs of Part III with the much more complex ones in this part. For instance, to represent the distributive lattice C1 +B2 as the congruence lattice of a ﬁnite sectionally complemented lattice, by Figure 7.3, we take the ideal lattice L of the chopped lattice MV , and we are done. The lattice L has 10 elements. Now start with the lattice K of Figure 12.1, a small seven-element lattice, whose congruence lattice is C1 + B2 . Even though K was chosen to minimize the size of the sectionally complemented congruence-preserving extension L, we ﬁnd this L much larger than the L of the representation theorem. In general, the L we construct is O(|K||K| ) in size. In [69], G. Gr¨ atzer and E. T. Schmidt proved the congruence-preserving extension theorem for sectionally complemented lattices. Theorem 12.1. Every ﬁnite lattice K has a ﬁnite, sectionally complemented, congruence-preserving extension L. The proof—as most proofs in Part IV—has two major ingredients: the representation theorem from Part III and the cubic extension from Part II, with a little bit of chopped lattices thrown in.

146

12. Sectionally Complemented Lattices

Since every ﬁnite sectionally complemented lattice is atomistic, this result contains the ﬁrst published result on congruence-preserving extension, namely, the theorem of Tischendorf [117]. Theorem 12.2. Every ﬁnite lattice has a ﬁnite atomistic congruence-preserving extension.

12.2.

Proof-by-Picture

Let us ﬁnd a ﬁnite, sectionally complemented, congruence-preserving extension for the lattice K of Figure 12.1. The ﬁgure also shows Con K, ConJ K, and ConM K; the congruence Φ1 splits K into two classes, marked by the dotted line (Φ2 is symmetric). Note that ConJ K ∼ = ConM K (Theorem 2.19). So K has three subdirectly irreducible quotients, K ∼ = K/ω, K/Φ1 ∼ = C2 , ∼ and K/Φ2 = C2 . To form a cubic extension (see Section 6), we embed the subdirectly irreducible quotients into simple lattices; C2 being simple, we only have to embed K. Figure 12.2 shows the Simp K we choose (compare it with the construct of Lemma 12.3), adding a single element a to K. So the cubic extension of K is Cube K = Simp K × C2 × C2 .

Φ1

ι

Φ3

Con K

Φ1 ω

Φ2 ω

Figure 12.1: The example lattice K.

a

K

Φ2 Φ3 ConJ K

Φ2

Φ1

K

Φ1

Simp K

Figure 12.2: A simple extension of K.

ConM K

12.2. Proof-by-Picture

a

sΦ1

147

sΦ2

Figure 12.3: The chopped lattice M .

This lattice is not easy to draw, but we easily draw the chopped lattice M of Figure 12.3, whose ideal lattice is Cube K. In this example, the boolean ideal is B—of Theorem 6.2—is generated by the atoms a, sΦ1 , and sΦ2 . Now recall Theorem 7.9: For D = Con K, there is a a ﬁnite sectionally complemented lattice L0 with Con L0 ∼ = D and with a boolean ideal B0 , the eight-element boolean lattice with atoms pΦ1 , pΦ2 , and pΦ3 so that con(0, pΦi ) represents Φi , for i = 1, 2, 3; see Figure 12.4. We merge Simp K and L0 by identifying the zeros and the atom a of Simp K with the atom of pΦ3 of L0 (and of B0 ), see Figure 12.5. We view M as being a part of the merged chopped lattice by identifying sΦ1 with pΦ† = pΦ2 and sΦ2 with pΦ† = pΦ1 ; Figure 12.5 shows the merged chopped 1 2 lattice, the black-ﬁlled elements form M . Basically, we identify the boolean ideal B of Cube K with the boolean ideal B0 of L0 . We deﬁne L as the ideal lattice of this chopped lattice. By the Atom

L0 B0 pΦ

3

pΦ

2

pΦ

1

Figure 12.4: The lattices L0 and B0 .

148

12. Sectionally Complemented Lattices

L0 B0

Simp K Figure 12.5: Merging L0 and M . Lemma—Lemma 4.8—L is a ﬁnite sectionally complemented lattice. The diagonal embedding Diag of K into Id M of Section 6.1 also embeds K into L, so (after identifying a ∈ K with Diag(a)) the lattice L can be regarded as an extension of K. What about the congruences of L? Clearly, merging L0 with a simple lattice, we obtain a chopped lattice that is a congruence-preserving extension of L0 , so L is a congruence-preserving extension of L0 and Con L ∼ = Con L0 ∼ = ∼ Con K = D. A congruence of L can be described as a compatible congruence vector Θ, Ψ of the chopped lattice, where Θ is a congruence of Cube K and Ψ is a congruence of L0 agreeing on {o, a} = {0, pΦ3 }. The congruence Θ is determined on B and the congruence Ψ is determined on B0 . In L, we identiﬁed B and B0 , so it follows that Θ = Ψ. Moreover, Θ, Θ is a compatible congruence vector, because Θ acts on B by collapsing pΦ and 0 iff Φ ≤ Θ, while Θ acts on B0 by collapsing pΦ† and 0 iff Φ† Θ; also, Φ ≤ Θ is equivalent to Φ† Θ. Since K has the congruence extension property in Cube K, we conclude that ϕ : Θ → con(Θ) (in L), for Θ ∈ Con K, is a bijection. On the other hand, Con L ∼ = Con L0 ∼ = Con K, so ϕ is an isomorphism between Con K and Con L, that is, L is a congruence-preserving extension of K.

12.3.

Simple extensions

To start out in the proof, we construct sectionally complemented cubic extensions. So we need the following statement. Lemma 12.3. For every ﬁnite lattice K, there is a ﬁnite, simple, sectionally complemented {0, 1}-extension L.

12.3. Simple extensions

149

Proof. We proceed by induction. Let 0 and 1 be the zero and unit element of L, respectively. If |K| ≤ 2, then we can take L = K. So we can assume that |K| > 2. We distinguish two cases. Case 1: 1 is join-irreducible. Let a be the unique dual atom of K. By the induction hypothesis, id(a) has a ﬁnite, simple, sectionally complemented {0, 1}-extension L . We deﬁne L = L ∪ {1, u, v}, where u and v common complements to all elements of L − {0} and are complementary, that is, u ∧ x = 0 and u ∨ x = 1, for all x ∈ (L ∪ {v}) − {0}; and symmetrically for v. It is easy to see that L satisﬁes all the requirements. Case 2: 1 is join-reducible. So there are elements x1 , x2 ∈ K − {1} satisfying x1 ∨ x2 = 1. Let N (K) = K − { x | x = 0 or x is an atom }. For every a ∈ N (K), we adjoin an atom pa < a so that if a = b, then pa = pb . We deﬁne on Simp K = K ∪ { pa | a ∈ N (K) } (a disjoint union) an ordering by the following rules: (α) K is a suborder of Simp K; (β) if a ∈ N (K) and x ∈ K, then x < pa iﬀ x = 0, pa < x iﬀ a ≤ x; (γ) if a and b ∈ N (K), then pa ≤ pb

iﬀ a = b.

It is easy to check that Simp K is a lattice. Obviously, Simp K is a ﬁnite lattice; 0 and 1 are the zero and unit elements of Simp K, respectively. Simp K is an extension of K. To show that Simp K is sectionally complemented, take 0 < u < v in Simp K. If u ∈ K, then v ∈ N (K) and pv is a sectional complement of u in [0, v]. If u ∈ / K, then u = pa , for some a ∈ N (K), and v ∈ K satisﬁes a ≤ v. If a = v, then any x satisfying 0 < x < a is a sectional complement of u in [0, v]; and there is such an x because a ∈ N (K). If a < v, then pv is a sectional complement of u in [0, v]. Finally, L is simple. Indeed, let Θ > ω be a congruence of K. We verify that there is an a > 0 in L, such that a ≡ 0 (Θ). Indeed, since Θ > ω, there are u, v ∈ K such that u < v and u ≡ v (Θ). If u = 0, then a = v satisﬁes the requirements. If 0 < u, then v ∈ N (K), so there is an element pv ∈ L.

150

12. Sectionally Complemented Lattices

If u ∈ K, then u ≡ v (Θ) implies that pv ≡ 0 (Θ), so we can take a = pv . If u ∈ Simp K − K, then u = px , for some x ∈ N (K). Obviously, v ∈ K and x ≤ v. Since x ∈ N (K), there is an a ∈ K satisfying 0 < a < x. Now u ≡ v (Θ) implies that px ≡ x (Θ) and so a ≡ 0 (Θ). Using the congruence a ≡ 0 (Θ), we conclude that p1 ≡ 1 (Θ). Meeting with x1 , we obtain that x1 ≡ 0 (Θ), and similarly x2 ≡ 0 (Θ). Joining these two congruences, we obtain that 0 ≡ 1 (Θ), that is, Θ = ι. We should point out that since a ﬁnite partition lattice is simple and sectionally complemented by Ore [99], Lemma 12.3 follows from the following very deep result of Pudl´ ak and T˚ uma [101]: Every ﬁnite lattice can be embedded into a ﬁnite partition lattice. It can also be derived from a much earlier result of R. P. Dilworth (ﬁrst published in Crawley and Dilworth [10]): Every ﬁnite lattice can be embedded into a ﬁnite geometric lattice.

12.4.

Formal proof

Let K be a ﬁnite lattice. In this section we construct a ﬁnite, sectionally complemented, congruence-preserving extension L of K, as required by Theorem 12.1. Using Lemma 12.3, for every Φ ∈ ConM K, we select a ﬁnite, sectionally complemented, simple extension KΦ = Simp K/Φ of K/Φ, with an atom, sΦ , and zero, 0Φ . Let D = Con K, and let L0 be a lattice whose existence was stated in Theorem 7.9 for the ﬁnite distributive lattice D. For a congruence Θ of K, let Θ0 denote the congruence of L0 determined by the set { qΦ | Φ ≤ Θ }, that is, Θ0 =

( con(qΦ , 0L0 ) | Φ ≤ Θ );

obviously, Θ → Θ0 sets up an isomorphism Con K → Con L0 . Now we inductively construct the lattice L of Theorem 12.1. Let Φ1 , . . . , Φn list the meet-irreducible congruences of K. We apply the Atom Lemma, see Lemma 4.8, with A = L0 , B = Simp K/Φ1 , and the atoms pΦ† of L0 and 1 sΦ1 of Simp K/Φ1 , to obtain the chopped lattice M1 and L1 = Id M1 . From Corollary 4.7, it follows that L1 is a congruence-preserving extension of L0 with the same boolean ideal B0 . Moreover, L1 is ﬁnite and sectionally complemented by the Atom Lemma. So we can apply again the Atom Lemma with A = L1 , B = Simp K/Φ2 , and the atoms pΦ† of L1 and sΦ2 of Simp K/Φ2 , 2 obtaining the chopped lattice M2 and L2 = Id M2 . In n steps, we construct Mn and L = Ln = Id Mn . It is clear that L is a ﬁnite sectionally complemented lattice. It is also evident that L is a congruence-preserving extension of L0 , hence, Con L ∼ = D.

12.4. Formal proof

151

So to complete the proof of Theorem 12.1, we have to show that K is congruence reﬂecting in L. In the next step, we need the following statement: Lemma 12.4. Let A, B, C be pairwise disjoint ﬁnite lattices with zeroes 0A , 0B , 0C , and atoms sA ∈ A, sB , sB ∈ B, sB = sB , sC ∈ C. We construct some chopped lattices by merging. (i) Let N1 be the chopped lattice obtained by forming the disjoint union of A and B and identifying 0A with 0B and sA with sB . (ii) Let N be the chopped lattice obtained by forming the disjoint union of A, B, C and identifying 0A with 0B and 0C , sA with sB , and sB with sC . (iii) Form the ideal lattice Id N1 , which we consider an extension of N1 , so it has atoms sA = sB and sB and let N2 be the chopped lattice that is the disjoint union of Id N1 and C with the zeroes identiﬁed and also the atom sB ∈ Id N1 is identiﬁed with the atom sC ∈ C. Then Id N2 is isomorphic to Id N . Proof. The elements of Id N1 are compatible pairs a, b ∈ A × B. The elements of Id N2 are compatible pairs a, b, c; note that a, b ∈ Id N1 iﬀ a ∧ sA = b ∧ sB and note also that a, b ∧ sB = b ∧ sB . On the other hand, the elements of Id N are compatible triples a, b, c ∈ A × B × C. It should now be obvious that a, b, c → a, b, c is the required isomorphism Id N2 → Id N . Another view of L is the following. We form the chopped lattice M1 ; instead of proceeding to L1 = Id M1 , let us now form by merging the chopped lattice M2 from M1 = M1 and Simp K/Φ2 by identifying the zeros of M1 and Simp K/Φ2 and the atom pΦ† of M1 with the atom sΦ2 of Simp K/Φ2 . Observe 2 that M2 is the union of L0 , Simp K/Φ1 , and Simp K/Φ2 and Simp K/Φ1 ∩ Simp K/Φ2 = {0}. By Lemma 12.4, Id M2 ∼ = L2 . Proceeding thus, we obtain the chopped lattice Mn = L0 ∪ Simp K/Φ1 ∪ Simp K/Φ2 ∪ · · · ∪ Simp K/Φn , where, for all 1 ≤ i < j ≤ n, we have Simp K/Φi ∩ Simp K/Φj = {0}, and L∼ = Id Mn , again by Lemma 12.4. So the chopped sublattice Simp K/Φ1 ∪ Simp K/Φ2 ∪ · · · ∪ Simp K/Φn of Mn is just the lattices Simp K/Φ1 , Simp K/Φ2 , . . . , Simp K/Φn merged by their zeros, and, therefore,

152

12. Sectionally Complemented Lattices

Id(Simp K/Φ1 ∪ Simp K/Φ2 ∪ · · · ∪ Simp K/Φn ) ∼ = Cube K. Thus we can view L as the ideal lattice of the chopped lattice obtained by gluing together L0 and Cube K over the boolean lattices B0 in L0 and B in Cube K. A congruence Θ of L is then determined by a congruence ΘL0 on L0 and a congruence ΘCube K on Cube K that agree on the boolean lattice. Since Cube K is a congruence-preserving extension of K ∼ = Diag(K), it follows now that L is a congruence-preserving extension of K, concluding the proof of Theorem 12.1.

12.5.

Discussion

Problem 12.1. By Lemma 12.3, every ﬁnite lattice K has a ﬁnite, simple, sectionally complemented extension L. What is the minium size of such an L? Problem 12.2. What is the size of the lattice L we construct for Theorem 12.1? What is the minimum size of a lattice L satisfying Theorem 12.1? Can we do better than Theorem 12.1? Problem 12.3. Is there a natural subclass S of SecComp for which Theorem 12.1 holds, that is, every ﬁnite lattice K has a ﬁnite, sectionally complemented, congruence-preserving extension L ∈ S. For inﬁnite lattices, almost nothing is known, as illustrated by the following question: Problem 12.4. Does every sectionally complemented lattice have a proper, sectionally complemented, congruence-preserving extension?

13 Semimodular Lattices

13.1.

The extension theorem

In Part III ﬁrst we considered ﬁnite sectionally complemented lattices, then ﬁnite semimodular lattices. We do the same in this part. In [72] we follow the same path with Schmidt as in Chapter 12. First, we have to ﬁnd a semimodular cubic extension; luckily, Crawley and Dilworth [10] comes to the rescue. Then we utilize the representation theorem for semimodular lattices, Theorem 9.1, to order the congruences. These two lattices have to be connected since there is no obvious way of amalgamating the two; we do this with the “conduit” construction. Thus we obtain the result of [72]: Theorem 13.1. Every ﬁnite lattice K has a congruence-preserving embedding into a ﬁnite semimodular lattice L.

13.2.

Proof-by-Picture

Let us start with a ﬁnite lattice K of Figure 13.1 (the dual of N6 , introduced in Section 6.1; see Figure 6.1), with Con K = {ω, Φ, ι}, the three-element chain. The congruence Φ splits K into two classes, marked in Figure 13.1 with the dashed line. First, we deﬁne a semimodular extension Cube K. To do this, we have to specify the function Simp. Since ConM K = {Φ, ω}, we have to deﬁne Simp K/Φ and Simp K/ω. The subdirectly irreducible quotient, K/Φ, is isomorphic to C2 , a simple semimodular lattice, so we can deﬁne Simp C2 = C2 . We embed the lattice K/ω ∼ = K into a simple semimodular lattice Simp K as in Figure 13.1. Therefore, the semimodular cubic extension,

154

13. Semimodular Lattices

Φ

Figure 13.1: The lattice K and a simple semimodular extension Simp K. Cube K, is deﬁned as C2 × Simp K. We thereby obtain a semimodular congruence-reflecting extension Cube K; see Figure 13.2. This ﬁgure also shows a “sketch” of the diagram, which we use in Figure 13.5 to visualize the gluings. The black-ﬁlled elements of Cube K in Figure 13.2 form the lattice K, identiﬁed with Diag(K). To see how the congruences of K extend to Cube K, we only have to describe the extension Cube(Φ) of Φ, which splits Cube K into two classes as shown by the dashed line in Figure 13.2. The shaded area in Figure 13.2 is (one option for) the ﬁlter F (of Corollary 6.3), which is a congruence-preserving sublattice of Cube K. We color the ﬁlter F with two colors, p1 and p2 . In Cube K, we have

p2

Cube(Φ)

p1

F p1

p2

p1 p2 p2 p1

Cube K

Figure 13.2: The semimodular cubic extension Cube K with sketch.

F

13.2. Proof-by-Picture

155

S p1 p2

p1 p2

p1 p1

p1 N

p2 p1

N

Figure 13.3: The lattice S with sketch. con(p1 ) = Cube(Φ). Also, Con(Cube K) = {ω, con(p1 ), con(p2 ), ι}. We now have to construct a congruence-determining extension of Cube K in which con(p1 ) = ι, so that con(p1 ) > con(p2 ). We borrow from Section 9.2 the semimodular lattice S; see Figure 13.3. We utilize the special circumstances to make S a little smaller. The lattice S contains an ideal N that is a chain of length 2; the two prime intervals colored by p1 and p2 . The ideal N is a congruence-determining sublattice of S. This ﬁgure also shows a sketch of the diagram. Note that in S, we have con(p1 ) > con(p2 ). We have to patch Cube K and S together, respecting the colors. We utilize for this the modular lattice M —which we call the ‘conduit’—shown in Figure 13.4 with a sketch. The lattice M has an ideal B, which is a

p2

p2

J

p1 J

p1 p1 p2

B

p2

B

p1 p2 p 2 p1

p1 Figure 13.4: The conduit with sketch. congruence-preserving sublattice isomorphic to F and a ﬁlter J, which is a congruence-preserving sublattice isomorphic to N . We color M as shown in Figure 13.4, so that there are color preserving isomorphisms ϕ : B → F and ψ : J → N. We glue Cube K and the conduit over the ﬁlter F and ideal B with respect to ϕ; the resulting lattice T has ﬁlter J. We glue T and S over the ﬁlter J and ideal N with respect to ψ to obtain the lattice L. The two gluings are

156

13. Semimodular Lattices

S S S

p2 p1

N p1

B

p2 p1

p2

N

J

p1 p2 p 2 p1

p1

M

p1

p1

J

p1 p2 p 2 p1

p1 p2 p 2 p1

T

F

p2

p2

p2

p2

T Cube K

p1

Cube K

Cube K

Figure 13.5: Two gluings: before and after; sketches. sketched in Figure 13.5. The semimodularity of L follows by two applications of Lemma 2.24. This semimodular congruence-preserving extension L of the six-element lattice K has 57 elements; compare this with the size of the lattice from Section 9.2: S has 24 elements.

13.3.

The conduit

In this section we give the general construction of the conduit, a modular lattice, which we constructed in a special case in Section 13.2; see Figure 13.4. Let n ≥ 1 be a natural number, and for every i, with 1 ≤ i ≤ n, we take a copy Ti of M3 , with atoms pi , qi , and ri . We form M3n = ( Ti | 1 ≤ i ≤ n ) with zero 0, and regard Ti as an ideal of M3n , so pi , qi , and ri are atoms of M3n , for 1 ≤ i ≤ n. n by { pi | 1 ≤ i ≤ n }; let 1B = Let B be the sublattice of M3 generated ( pi | 1 ≤ i ≤ n ). Obviously, B is a 2n -element boolean lattice, an ideal of M3n .

13.4. The construction

Deﬁne qi =

157

( qj | 1 ≤ j ≤ i ),

for 1 ≤ i ≤ n, and set E = { q i | 0 ≤ i ≤ n }, where q 0 is the zero of M3n . Obviously, E is a maximal chain (of length n) in id(q n ) of M3n and b ∧ e = 0, for all b ∈ B and e ∈ E. By Theorem 2.11, the sublattice A of M3n generated by B and E is isomorphic to B × E under the isomorphism b ∨ e → b, e. Let b ∈ B and let i, with 0 ≤ i < n, satisfy the condition: pi+1 b. Deﬁne the element of M3n : r(b, i) = b ∨ q i ∨ ri+1 , and the subset M of M3n : M = A ∪ { r(b, i) | b ∈ B, 0 ≤ i < n, pi+1 b }. M is a sublattice of M3n , hence M is a modular lattice. The lattice M contains B and E as ideals. Let Θ be a congruence of B. Let ΘE be the congruence on E satisfying q i ≡ q i+1 (ΘE ) in E iﬀ pi+1 ≡ 0 (Θ) in B, for 0 ≤ i < n. Then Θ × ΘE is a congruence on B × E. We extend this to a congruence ΘM of M as follows: let r(b, i) be deﬁned (that is, b ∈ B and pi+1 b); if b ≡ b ∨ pi+1 (Θ) in B, then r(b, i) ∈ (b ∨ q i )/ΘM , otherwise, {r(b, i)} is a congruence class. The map Θ → ΘM is an isomorphism between Con B and Con M . In fact, M is a congruence-preserving extension of both B and E. Let J = { 1B ∨ e | e ∈ E }. Obviously, E and J are isomorphic chains and J is a ﬁlter of M , a congruencepreserving sublattice of M . To summarize: Lemma 13.2. For each n ≥ 1, M3n has sublattices B, J, and M satisfying the following conditions: (i) B is an ideal of M and it is isomorphic to the boolean lattice Bn . (ii) J is a ﬁlter of M and length J = n. (iii) M is a congruence-preserving extension of both B and J. Note that Con M is a boolean lattice and Con M ∼ = Con J. = Con B ∼

13.4.

The construction

We are given a ﬁnite lattice K. To prove the Theorem 13.1, we have to construct a ﬁnite, semimodular, congruence-preserving extension L of K.

158

13. Semimodular Lattices

We glue L together from three lattices, sketched in Figure 13.5. The ﬁrst lattice is S of Section 9.3 (denoted there by L)—see Figure 9.4—with the ideal N , deﬁned at the end of Section 9.3, which is a chain consisting of the gray-ﬁlled elements, a congruence-determining sublattice of S. Let D = Con K; in Section 9.3, we construct a ﬁnite semimodular lattice S with D ∼ = Con S. The construction of S in Section 9.3 starts out by representing D as Down P , for P = J(D). By Theorem 2.19, the orders J(D) and M(D) are isomorphic, so we can start the construction of S with P = M(D). Accordingly, we color N with P = M(D); there is a one-to-one correspondence between the prime intervals of N and the elements of P , the meet-irreducible congruences of K. The second lattice is the conduit lattice M of Section 13.3, with the ideal B (which is boolean) and the ﬁlter J (which is a chain), constructed so that J and N are isomorphic. This (unique) isomorphism allows us to color J so that the isomorphism preserves colors. The coloring of J extends to a coloring of all of M , in particular, to a coloring of B. Both B and J are congruence-preserving sublattices of M . The third lattice is a cubic extension, so we have to deﬁne the function Simp. For a ﬁnite lattice A, let Simp A be a ﬁnite partition lattice extending A, a semimodular lattice by Lemma 2.25. Such an extension exists by Pudl´ ak and J. T˚ uma [101]—alternatively, use the more accessible result of Crawley and Dilworth [10] (Theorem 14.1) to obtain a simple semimodular extension. In the cubic extension Cube K, we have the ﬁlter F generated by dual atom tΨ ∈ Simp KΨ , for Ψ ∈ ConM K = P , by Corollary 6.3. We color [tΨ , 1F ] by Ψ, for Ψ ∈ ConM K = P , and thereby color all of F . Observe that B and F are isomorphic ﬁnite boolean lattices, both are colored by P , so there is an isomorphism (unique!) α between B and F that preserves colors. First gluing. We glue M and S together over J and N to obtain the lattice T . The lattice T is also colored by P and T has an ideal B, a congruence-determining sublattice. The lattice T is obtained by gluing together two semimodular lattices, hence it is semimodular by Lemma 2.24. Second gluing. We have the color preserving isomorphism between the boolean ideal B of the lattice T of the ﬁrst step and the ﬁlter F in Cube K; we do the second gluing of T and B with Cube K and F with respect to this isomorphism α, to obtain the lattice L. The lattice L is obtained by gluing together two semimodular lattices, hence it is semimodular by Lemma 2.24.

13.5. Formal proof

13.5.

159

Formal proof

To prove Theorem 13.1, we have to verify that L is a ﬁnite, semimodular, congruence-preserving extension of K. We already know that L is ﬁnite and semimodular. We view K as a sublattice of Cube K (as in Section 6.1), so K is a sublattice of L. Let Ω be a congruence of K. We have to show that Ω has one and only one extension to L. Let Ψ ∈ ConM K be the color of the prime interval p of L. Let qΨ be the unique prime interval in N of color Ψ. We claim that con(p) = con(qΨ ). To prove this claim, let p be a prime interval in L. Then p is a prime interval in S, or M , or Cube K. So we have three cases to consider. First, if p is a prime interval in S, this follows from the fact that N is a congruence-determining sublattice of S; see Lemma 3.18 and the last paragraph of Section 9.3. Second, if p is a prime interval in M , this follows from the fact that J is a congruence-preserving sublattice of M , by Lemma 13.2. Third, if p is a prime interval in Cube K, then there is a prime interval p1 in F with con(p) = con(p1 ), since F is a congruence-determining sublattice of Cube K. Because of the gluing, the prime interval p1 can be regarded as a prime interval of M , and now the claim follows from the second case. This completes the proof of the claim. It follows from this claim that a congruence Θ of L can be described by the set Σ of prime intervals of N collapsed by Θ, or equivalently, by the set of colors of the prime intervals: C = { Ψ | pΨ ∈ Σ }. Now by the construction of S (see Section 9.3), the set C is a down-set of P = ConM K. And conversely, if C is a down-set of P , then it deﬁnes a congruence ΘC on S, as computed in Section 9.3. This congruence ΘC spreads uniquely to M because J is a congruence-preserving sublattice of M . Finally, from the description of Cube Ω in Section 6.2, we see that the congruence uniquely spreads to Cube K. So there is a one-to-one correspondence between congruences of L and down-sets of P . For Ω ∈ Con K, its extension Cube(Ω) to Cube K corresponds to the down-set { Ψ | Ω Ψ }. So every congruence of K extends. But since there are no more congruences than those that correspond to down-sets of P , the extension must be unique. This completes the proof of Theorem 13.1.

13.6.

Discussion

Using Crawley and Dilworth [10], we get a very large simple semimodular lattice extending the lattice K. Of course, from Pudl´ ak and T˚ uma [101] we get an even larger lattice. So it is natural to ask:

160

13. Semimodular Lattices

Problem 13.1. Let f (n) be the smallest integer with the property that every lattice K of n elements can be embedded into a simple semimodular lattice of size f (n). Give an asymptotic estimate of f (n). Problem 13.2. What is the size of the lattice L we construct for Theorem 13.1 expressed in terms of f (n) of Problem 13.1? What is the minimum size of a lattice L satisfying Theorem 13.1? Can we do better than Theorem 13.1? Problem 13.3. Is there a natural subclass S of the class of semimodular lattices for which Theorem 13.1 holds, that is, every ﬁnite lattice K has a ﬁnite, semimodular, congruence-preserving extension L ∈ S. How about planar lattices? Problem 13.4. Does every planar lattice K have a congruence-preserving embedding into a ﬁnite, planar, semimodular lattice L?

14 Isoform Lattices

14.1.

The result

In Chapter 11, we proved the representation theorem for ﬁnite uniform lattices. In Section 11.5, we discussed the representation theorem for ﬁnite isoform lattices, but we did not prove it, for a reason I am about to explain. In Gr¨ atzer, Quackenbush, and Schmidt [53], we proved the following: Theorem 14.1. Every ﬁnite lattice K has a congruence-preserving extension to a ﬁnite isoform lattice L. This problem was raised for uniform lattices as Problem 1 in Gr¨ atzer, Schmidt, and Thomsen [79] and for isoform lattices as Problem 2 in Gr¨ atzer and Schmidt [76]. (See also Problem 9 in Gr¨ atzer and Schmidt [78].) The proof of this theorem breaks the earlier pattern: we start constructing a congruence-preserving extension by a cubic extension and then we utilize the representation theorem. For isoform lattices, we again start with a cubic extension, but we do not know how to proceed utilizing the representation theorem—this is why it was not important to have the representation theorem proved in Chapter 11. Instead, we use “pruning.” Special cases of pruning are the A+ construction of Section 11.2, the N (A, B) construction of Section 11.3, and the N (A, B, Θ) construction of Section 11.5.

14.2.

Proof-by-Picture

In this section we construct the lattice L of Theorem 14.1 for the lattice K = N6 of Figure 14.1.

162

14. Isoform Lattices

1

Φ d v

a b

c

0 Figure 14.1: The lattices K and Sω . As always, when we construct a congruence-preserving extension, we start with a cubic extension. To construct the cubic extension of K, we have to deﬁne the function Simp on the subdirectly irreducible quotients. The lattice K has only one nontrivial congruence Φ, splitting K into two classes, indicated by the dashed line; it has two meet-irreducible congruences, Φ and ω. We use the notation: Simp K/ω = Sω and Simp K/Φ = SΦ . The choice for Sω is easy; add a single element v as in Figure 14.1 (the black-ﬁlled element). The lattice K/Φ is simple, so we can choose SΦ = C2 .

1 1ω

1Φ 0 Figure 14.2: The cubic extension S of K. Figure 14.2 shows the cubic extension S = Sω × SΦ of K; the black-ﬁlled elements represent K. On the diagram, we labeled four elements: the 0 and 1 of S, and the unit elements of the two direct factors: 1ω and 1Φ . The congruences of S are in one-to-one correspondence with subsets of {1ω , 1Φ }, and Cube Φ (the extension of Φ to S) corresponds to {1ω }, that is, Cube Φ = con(0, 1ω ). So we have to “kill” the congruence con(0, 1Φ ); in other words, we have to make sure that 1Φ ≡ 0 implies that 1ω ≡ 0.

14.2. Proof-by-Picture

163

In Chapters 12 and 13, such a congruence-forcing was accomplished by a suitable extension of S utilizing the appropriate representation theorems. The inspiration for the isoform result comes from the representation theorem: the use of pruning, although the representation theorem itself is not needed. We construct L from S by pruning (deleting) a single edge, the dashed edge of Figure 14.2; the resulting order L is depicted in Figure 14.3.

Figure 14.3: The lattice L. It has to be checked that L is a lattice (pruning an edge does not, in general, produce a lattice). Then we have to see that 1Φ ≡ 0 implies that 1ω ≡ 0 in L, so L has only one nontrivial congruence and it splits L into two classes as indicated by the dashed line. The two classes are both isomorphic to Sω , so L is isoform. Computing this example in detail, one comes to the conclusion that everything works out because the element v in Sω is so special. We axiomatize the important property of v as follows. For a ﬁnite lattice A, let us call an element v a separator if 0 ≺ v ≺ 1. A lattice A with a separator is called separable; we also allow C2 as a separable lattice. The general construction starts with a ﬁnite order P (which plays the role of ConM K), and a family Sp , for p ∈ P , of separable lattices. We form the set S = ( Sp | p ∈ P ). In each Sp = C2 , with p ∈ P , we ﬁx a separator vp . Let us describe the edges we delete from S. Let a = ap p∈P ≺ b = bp p∈P in S. Then there is exactly one q ∈ P with aq ≺ bq in Sq , and ap = bp , for all p = q. We delete the edge a, b if there is a p < q such that Sp has a separating element vp and vp = ap = bp . The pruned diagram deﬁnes the order L. Here is a small example. The order is P = {p, q} with p < q and the two lattices are Sp = C3 and Sq = C2 . Figure 14.4 illustrates the construction; the edge of S = C3 × C2 to be pruned is dashed.

164

14. Isoform Lattices

1, 1

1, 0

v, 1

1, 0 v, 0

S

1, 1

p
0, 1

Sp

Sq

v, 1 0, 1

v, 0

L

0, 0

0, 0

Figure 14.4: A small example of the construction of L.

In Figure 14.5, we illustrate the construction with the order P = {p, q, r} (the hat) with p < q and r < q, with the lattices Sp = Sq = Sr = C3 = {0, v, 1}. Four edges of C33 are missing in L; on the diagram, these are marked with dashed lines. It is not so obvious that the order L deﬁned by the pruned diagram is a lattice.

r

uP p

q

P

u{p,r}

u{p,q}

up

u{q,r}

v, 1, 0 0, 1, v

v, v, 0 v, 0, 0

Sp

uq

0, v, v 0, 0, v

Sq

Sr

u∅ Figure 14.5: A nontrivial example of the construction.

ur

14.3. Formal construction

165

In Sections 14.4 and 14.5 we have to assume that each Sp is simple and Sp = C2 . So even the smallest appropriate example is fairly large. To construct the smallest example, let P = {p, q} with p < q and Sp = Sq = M3 = {0, a, b, v, 1}. Figure 14.6 illustrates what we get; note that the diagram is turned to its side. In this diagram, Sq is on the right, its elements are labeled 0, a, b, v, 1; its unit element is uq = 0, 1. The lattice Sp is on the left, its elements are all labeled with 0; its unit element is up = 1, 0. Five edges of M32 are missing in L; on the diagram these are marked with dashed lines. It is easy to see that the order L deﬁned by the pruned diagram is a lattice. This is how far we can get with pictures. We can illustrate the basic construction, but have no Proof-by-Picture that L is a lattice. This would have to be computed; we have no special insight why the computation works. It is even worse to verify the isoform property . . .

14.3.

Formal construction

If there is no ambiguity, the subscripts will be dropped. In this section for a ﬁnite order P , and a family Sp , for p ∈ P , of separable lattices, we form the set S = ( Sp | p ∈ P ). In S, ∨S , ∧S , ≤S , ≺S denotes the join, meet, ordering relation, and covering relation, respectively. If there is no ambiguity, the subscript for v will be dropped. The elements of S will be bold face lower case letters. An element s ∈ S is of the form sp p∈P , and we write sp for sp . For q ∈ P , deﬁne the elements uq ∈ S and vq ∈ S as follows: 1, (uq )p = 0, v, (vq )p = 0,

if p = q; if p = q; if p = q; if p = q.

Let B be the sublattice of S generated by { up | p ∈ P }. We call B the skeleton of S; itis a boolean sublattice of S with n atoms. For a subset Q of P , set uQ = S {up | p ∈ Q} with complement (uQ ) = uP −Q in B (and in S). The elements of B are black-ﬁlled in Figure 14.5. Now we come to the crucial deﬁnition of this chapter. We deﬁne a binary relation ≤ on the set S. For the elements a = ap p∈P and b = bp p∈P of S: (P) a ≤ b iff a ≤S b and p < p in P, Sp = C2 , and ap = vp = bp

imply that ap = bp .

In this section and in the next, when we write “p ∈ P, and ap = vp = bp ”,

Sq 0, 0

b, 0

a, 0

Sp

0

b

v, 0 0

v, v

1, 0

up

0

v

v

v, a

v, b

1 a

a

v, 1

b

1

1, 1

v

a

1

b

0

v

0

a

v

1

b

p
a

1

b

0, 1

14. Isoform Lattices

uq

166

Figure 14.6: The smallest example of the construction with simple lattices of more than two elements—sideways.

14.3. Formal construction

167

we will mean implicitly that Sp = C2 (so that vp be deﬁned). In Sections 14.4 and 14.5 we assume that each Sp is simple and Sp = C2 , so this implicit assumption always holds, for all p ∈P. Let L be the relational system S, ≤. In Theorem 14.2, we show that L is an order and in Theorem 14.4, we verify that L is a lattice. Note that if P is an antichain, then ≤ is the same as ≤S , so L = S as a relational system. Figure 14.4 illustrates the smallest example where S and L diﬀer. Now we proceed to prove that L is a lattice and verify some properties of the covering relation in L. Theorem 14.2. L is an order. Proof. The relation ≤ is obviously reﬂexive. By the deﬁnition of ≤: If a ≤ b, then a ≤S b. So the anti-symmetry of ≤S implies the anti-symmetry of ≤. To prove the transitivity of ≤, let a, b, c ∈ L, and let a ≤ b ≤ c. Then a ≤S b ≤S c, so by the transitivity of ≤S , we get that a ≤S c. To verify (P) for a and c, let p < p in P , and assume that ap = v = cp (recall our implicit assumption: Sp = C2 ). Since a ≤S b ≤S c, it follows that ap ≤ bp ≤ cp , and we conclude that ap = bp = cp = v. Since a ≤ b and b ≤ c, applying (P) twice, we obtain the equalities ap = bp , and bp = cp . It follows that ap = cp , verifying (P) for a and c, hence proving that a ≤ c. Therefore, ≤ is transitive. We will need the following property of the order L: Lemma 14.3. (1) If a ≺ b in L, then a ≺S b in S. (2) If a ≺S b in S, then a ≺ b in L unless there are q < q in P with aq = v = bq and aq ≺ bq , in which case, a b in L. Proof of (1). We start by observing that a ≺S b iff there is a unique q ∈ P such that aq ≺ bq in Sq , and ar = br , for all r ∈ P , otherwise. Let us assume that a ≺ b in L. Then a <S b, by (P). Thus there is q ∈ P such that aq < bq . We claim that aq ≺ bq in Sq . Indeed, if aq < c < bq , for some element c ∈ Sq , then deﬁne c = cp p∈P as follows: ⎧ ⎪ ⎨c, if p = q; cp = v, if ap = v = bp ; ⎪ ⎩ x, with x ∈ {ap , bp } − {v}, otherwise.

168

14. Isoform Lattices

Obviously, a <S c <S b. We now verify that a < c. Let p < p in P and let ap = cp = v. Notice that ap = cp = bp = v. Since a < b, by (P), it follows that ap = bp . But then ap = cp , verifying a < c by (P). We can verify that c < b, similarly. Therefore, a < c < b, contradicting the assumption that a ≺ b. This proves the claim. Thus we may assume that aq ≺ bq . We now claim that there is only one q with aq < bq . Indeed, let us assume that there are q = r in P with aq < bq and ar < br . We assume that q is minimal in P with the property: aq < bq . There are two cases to consider. Case 1: bq = v. Since v is a separator, it follows that aq = 0. Deﬁne c = cp p∈P by ap , for p = q; cp = bp , for p = q. Obviously, a <S c <S b. Since aq = 0 and cp = bp , for p = q, therefore, cp = v iff bp = v. We now verify that a < c. Let p < p in P and let ap = cp = v. We cannot have p = q, because aq = 0. Thus p = q, and so ap = bp = cp = v. Now (P) and a < b imply that ap = bp . For any p , either ap or bp equals cp . Therefore, ap = bp = cp , concluding that a < c. We next verify that c < b. Let p < p in P and let cp = bp = v and assume to the contrary that cp < bp . By the deﬁnition of c, this is possible only if p = q, in which case the minimality of q forces ap = bp = v, contradicting a < b. Thus c < b. We conclude that a < c < b, contradicting that a ≺ b. Case 2: bq = v. Deﬁne c = bp p∈P by

bp , for p = q; cp = ap , for p = q.

Obviously, a <S c <S b. We now verify that a < c. Let p < p in P and let ap = cp = v. Assume, to the contrary, that ap = cp . By the deﬁnition of c, then p = q. Since q is minimal with respect to aq < bq , it follows from p < p = q that ap = bp . So ap = bp = v and ap < bp , which by (P) contradicts the assumption a < b. This proves that a < c. We next verify that c < b. Let p < p in P and let cp = bp = v. By the assumption of Case 2, bq = v, and so p = q. We conclude that cp = bp = ap = v. Since a < b, by (P), we obtain that ap = bp . Therefore, p = q. By the deﬁnition of c, we have that ap = cp , so ap = bp = cp , verifying c < b.

14.3. Formal construction

169

We conclude that a < c < b, contradicting that a ≺ b. This completes the proof of (1). Proof of (2). Now assume that a ≺S b. Then there is a unique q ∈ P with aq ≺ bq , and otherwise, ar = br , for r ∈ P . Consequently, if a < b, then a ≺ b in L. If a < b fails, then by (P), there are elements p < p in P with ap = v = bp and ap < bp , implying that a b in L. Now we prove that we have constructed a lattice and describe the lattice operations. To facilitate this, we introduce the following terminology: Let a = ap p∈P , b = bp p∈P ∈ L, and let q ∈ P ; we will call the element q an {a, b}-fork if aq = bq = v and aq = bq , for some q > q. Theorem 14.4. L is a lattice. Let a = ap p∈P , b = bp p∈P ∈ L. Then ⎧ ⎪ 1, if ap ∨ bp = v and, for some p ≥ p, ⎪ ⎪ ⎪ ⎪ ⎪ (1∨ ) p is an {a, b}-fork, or ⎨ (a ∨ b)p = (2∨ ) bp ≤ ap and bp ap , or ⎪ ⎪ ⎪ (3∨ ) ap ≤ bp and ap bp ; ⎪ ⎪ ⎪ ⎩a ∨ b , otherwise; p p and

(a ∧ b)p =

⎧ ⎪ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩a ∧ b , p p

if ap ∧ bp = v and for some p ≥ p, (1∧ ) p is an {a, b}-fork, or (2∧ ) bp ≥ ap and bp ap , or (3∧ ) ap ≥ bp and ap bp ; otherwise.

Proof. Recall that v is separable, and therefore join-irreducible, hence ap ∨ bp = v is equivalent to ap , bp ∈ {0, v, v, 0, v, v}; and similarly, ap ∧bp = v is equivalent to ap , bp ∈ {1, v, v, 1, v, v}. Also notice that if ap = bp = v and Case (1∨ ) occurs, then p is an {a, b}-fork; in Cases (2∨ ) and (3∨ ), we must have p > p. Let c be the element deﬁned in the join formula. To see that a ≤ c (using the deﬁnition (P) of ≤), ﬁrst note that a ≤S a ∨S b ≤S c, so that a ≤S c. Let ap = cp = v and p > p; we must show that ap = cp . Now ap = cp = v implies that ap = ap ∨ bp = cp = v, and so bp ≤ ap = v. Hence, for all q ≥ p, none of Cases (1∨ )–(3∨ ) occur. That is, for all q ≥ p, q is not an {a, b}-fork (Case (1∨ )) and bq ≤ aq (Case (2∨ )). In Case (3∨ ), using that bp ≤ ap = v if ap ≤ bp , then ap = bp = v, and since p is not an {a, b}-fork, aq = bq , for q ≥ p. Consequently, bp ≤ ap . Hence, ap = ap ∨bp ≤ cp , with equality holding unless v = ap = ap ∨ bp and one of Cases (1∨ )–(3∨ ) holds, for some p ≥ p . But Case (1∨ ) cannot occur because p = q ≥ p is not an {a, b}-fork, and Case (2∨ ) cannot occur since p ≥ p forces bp ≤ ap . For Case (3∨ ), note

170

14. Isoform Lattices

that bp ≤ ap , since p > p. Hence, ap ≤ bp = v forces ap = bp = v. But p > p is not an {a, b}-fork, so ap = bp , and so Case (3∨ ) cannot occur. Thus ap = cp and a ≤ c. Similarly, b ≤ c. It remains to show that c is the least upper bound of a, b in L. So let a, b ≤ d. Then a, b ≤S d, so a ∨S b ≤S d. We ﬁrst show that c ≤S d. For p ∈ P if ap ∨ bp = cp , then cp ≤ dp . Let ap ∨ bp < cp ; as in the previous paragraph, it follows that ap ∨ bp = v < 1 = cp . Therefore, v ≤ dp ≤ 1. If dp = 1, then cp ≤ dp , as required. Now assume that dp = v. Since ap ∨ bp = v < 1 = cp , one of Cases (1∨ )–(3∨ ) must occur; in each case, we will arrive at a contradiction. If Case (1∨ ) occurs, then ap = bp = dp = v and, for some p > p, we have ap = bp . We conclude that dp cannot equal both ap and bp , contradicting that d is a common upper bound of a and b. If Case (2∨ ) occurs, then bp ≤ ap = v = dp and bp ap , for some p > p. Then dp ≥ ap ∨ bp > ap , contradicting that a ≤ d. Symmetrically, Case (3∨ ) leads to a contradiction. Thus c ≤S d. To prove that c ≤ d, let cp = dp = v and choose p < p ; we must show that cp = dp . Since cp = dp = v, the join formula tells us that ap ∨ bp = cp = v; without loss of generality, we may assume that ap = v so that ap = v = dp . As c ≤S d, either cp = dp or cp < dp . In the latter case, we have ap ≤ cp < dp , contradicting that a ≤ d. Hence, cp = dp . This proves that c ≤ d and therefore c is the least upper bound of a, b in L. The proof of the meet formula is similar, mutatis mutandis. From now on, for a, b ∈ L, a ≤ b, a ∨ b, a ∧ b, refer to the ordering and operations in L. We will need two lemmas that immediately follow from the join and meet formulas. For p ∈ P , the element v is doubly irreducible in Sp . Therefore, each Sp = Sp − {v} is a sublattice of Sp . Let S be the direct product of the Sp , p ∈ P . Obviously, S is a sublattice of S. Lemma 14.5. (1) S is a sublattice of L. (2) For p ∈ P , the interval [0, up ] is a sublattice of both L and S; in fact, the same sublattice, that is, (≤)[0,up ] = (≤S )[0,up ] . Proof. The proof follows from the join and meet formulas in Theorem 14.4. Lemma 14.6. Let a ∈ L and let D be a down-set of P . Then a ∨ uD = a ∨S uD , and

a ∧ (uD ) = a ∧S (uD ) .

14.4. The congruences

171

Proof. By the join formula, (a ∨ uD )p = (a ∨S uD )p unless ap = v, p ∈ / D, and ap < 1, for some p < p with p ∈ D. But this cannot happen since D is a down-set. By the meet formula, (a ∧ (uD ) )p = (a ∧S (uD ) )p unless ap = v, p∈ / D, and ap > 0, for some p < p with p ∈ D. But this cannot happen either since D is a down-set.

14.4.

The congruences

Now let each Sp be simple and let each Sp = C2 ; then S has a congruence lattice isomorphic to B such that uQ ∈ B is associated with the smallest congruence on S collapsing 0 and uQ , denoted ΘQ . Clearly, S is isoform. We are going to prove that the congruences of L are just the congruences ΘD of S for D a down-set of P . As an illustration, in Figure 14.6, the lattice has exactly one nontrivial congruence, “projecting” L onto Mq . The ﬁve congruence classes have ﬁve elements each, labeled x, with x ∈ {0, v, a, b, 1}. For p ∈ P , the lattice Sp is assumed to be simple, so the set Sp −{0, 1, v} = ∅. Therefore, we can select wp ∈ Sp − {0, 1, v}; we will write w for wp if the index is understood. We deﬁne wq by (wq )q = w and otherwise, (wq )p = 0. Lemma 14.7. Let Θ be a congruence of L, and let 0 ≡ up (Θ), for some p ∈ P . If q < p, then 0 ≡ uq (Θ). Proof. From 0 ≡ up (Θ) and q < p, the join formula yields the congruence vq = 0 ∨ vq ≡ up ∨ vq = u{p,q} (Θ). Similarly, we get the congruence 0 = wq ∧ vq ≡ wq ∧ u{p,q} = wq (Θ). Lemma 14.5 implies that [0, uq ] is a simple sublattice of L, so we obtain that 0 ≡ uq (Θ). Lemma 14.8. Let Θ be a congruence of L and D a down-set of P . Let a, b ∈ L with ap = 0 and bp = 1, for p ∈ D and ap = bp , for p ∈ / D. Then a ≤ b and 0 ≡ uD (Θ) iff a ≡ b (Θ). Proof. Clearly, b = a ∨S uD = a ∨ uD , by Lemma 14.5. Thus a ≤ b. Let 0 ≡ uD (Θ). Then a = 0 ∨ a ≡ uD ∨ a = b (Θ). Conversely, if a ≡ b (Θ), then 0 = a ∧ uD ≡ b ∧ uD = uD (Θ). Lemma 14.9. Let Θ be a congruence of L and let a ≺ b in L with a ≡ b (Θ). Then there is a unique p ∈ P with ap ≺ bp and aq = bq , for p = q; moreover, 0 ≡ up (Θ). Proof. If a ≺ b, then from Lemma 14.3, there is a unique p ∈ P with ap ≺ bp and aq = bq otherwise. Moreover, if aq = bq = v, then q p. If bp = v so that ap = 0, then we get (a ∨ wp )p = w < 1 = (b ∨ wp )p , and otherwise (a ∨ wp )q = aq = bq = (b ∨ wp )q since aq = bq = v implies that q p, so that (wp )r = 0, for all q < r. This yields the covering a ∨ wp ≺ b ∨ wp and the congruence a ∨ wp ≡ b ∨ wp (Θ). Thus we may assume that bp = v. Then 0 ≤ a ∧ up < b ∧ up ≤ up and a ∧ up ≡ b ∧ up (Θ). By Lemma 14.5, the interval [0, up ] is a simple sublattice of L. Hence, 0 ≡ up (Θ).

172

14. Isoform Lattices

Theorem 14.10. Let Θ be a congruence of L; then there is a down-set D of P such that Θ = ΘD . Conversely, let D be a down-set of P ; then ΘD is a congruence of L. Proof. Let Θ be a congruence of L. Deﬁne Prec(Θ) to be the set of all a, b ∈ L2 such that a ≡ b (Θ) and a ≺ b. Deﬁne D to be the set of all p ∈ P such that ap ≺ bp for some a, b ∈ Prec(Θ). By Lemma 14.9, 0 ≡ up (Θ), for all p ∈ D. By Lemma 14.7, D is a down-set of P . By Lemma 14.5, ( up | p ∈ D ) = uD ; hence, 0 ≡ uD (Θ). By Lemma 14.8, ΘD ⊆ Θ. On the other hand, since L is ﬁnite, Θ is the smallest equivalence relation containing Prec(Θ), so we must have Θ ⊆ ΘD . Thus Θ = ΘD . Conversely, let D be a down-set of P . A typical ΘD -class is of the form [a, b], where for p ∈ D, ap = 0 and bp = 1, and otherwise aq = bq . By Lemma 14.6, a ≤ a ∨ uD = a ∨S uD = b. Let c ∈ L; it suﬃces to show that a ∨ c ≡ b ∨ c (Θ) and a ∧ c ≡ b ∧ c (Θ). For the join, we may take a ≤ c, so that a ∨ c = c ≤ b ∨ c. Let us assume that cp < (b∨c)p ; we must show that p ∈ D. If cp < bp ∨cp , then from ap ≤ cp , we conclude that ap = bp , and so p ∈ D. Otherwise, cp = bp ∨ cp < (b ∨ c)p so that cp = v = bp ∨ cp < 1 = (b ∨ c)p . Again, if ap = bp , then p ∈ D; so assume that ap = bp ≤ cp = v. As D is a down-set, we must have aq = bq for all q ≥ p. Now, bp ∨ cp = v < 1 = (b ∨ c)p implies that one of Cases (1∨ )–(3∨ ) holds for {b, c}, for some p ≥ p. Note that bp ≤ cp implies that Case (3∨ ) cannot occur. If Case (1∨ ) occurs, then ap = bp = cp = v; since p is a {b, c}-fork, this forces p to be an {a, c}-fork, contradicting that a ≤ c. If Case (2∨ ) occurs, then ap = bp cp , again contradicting a ≤ c. This proves that a ∨ c ≡ b ∨ c (Θ). For the meet, we proceed similarly.

14.5.

The isoform property

As in Section 14.4, we assume that each Sp is simple and Sp = C2 . Theorem 14.11. L is isoform. Proof. Let D be a down-set of P ; in S every block of ΘD can be written in the form [a, a ∨S uD ], for some a ≤S (uD ) . Hence if b, c ∈ [a, a ∨S uD ] and bp = cp , then p ∈ D. The map φa : [0, uD ] → [a, a ∨S uD ] deﬁned by φa (x) = x ∨S uD is an isomorphism of sublattices of S. By Lemma 14.6, a ≤ a ∨ uD = a ∨S uD ; thus the blocks of ΘD in L are also of the form [a, a ∨S uD ] for some a ≤ (uD ) (which is the same as a ≤S (uD ) ). We will show that φa is also an isomorphism of L-sublattices; this will prove that L is isoform. We will make use without further mention that for b, c ∈ L, (b ∨S c)p = bp ∨ cp , for all p ∈ P . Obviously, φa is a bijection. Let us assume that 0 ≤ b ≺ c ≤ uD . Then by Lemma 14.6 and the fact that φa is an S-isomorphism, it follows that

14.6. Discussion

173

b ∨S a ≺S c ∨S a. If we do not have b ∨S a ≺ c ∨S a, then by Lemma 14.3, there are p < p with bp ∨ ap = v = cp ∨ ap and bp ∨ ap ≺ cp ∨ ap . From this latter, we get p ∈ D; since D is a down-set and p < p , we have p ∈ D. But then a ≤ (uD ) implies that ap = ap = 0; therefore, v = bp ∨ ap = bp and v = cp ∨ ap = cp , so that bp = cp = v. Since b ≺ c, we have that bp = cp for all p < p . For p = p , this yields bp ∨ ap = bp = cp = cp ∨ ap , contradicting that bp ∨ ap ≺ cp ∨ ap . Thus b ∨S a ≺ c ∨S a. Now assume that a ≤ b ≺ c ≤ a∨S uD ; hence, a ≤S b ≺S c ≤S a∨S uD . Since ϕa is an S-isomorphism, there are unique 0 ≤S b ≺S c ≤S uD such that b = b ∨S a and c = c ∨S a. Hence there is a unique p ∈ D such that bp ≺ cp , bp ∨ ap ≺ cp ∨ ap , and bq ∨ aq = cq ∨ aq , for p = q. We need to show that b ≤ c. So let br = v = cr ; then r = p. We have to verify that bq = cq , for r < q. This can fail only if r < p. But then r < p ∈ D, a downset; hence, r ∈ D. Since a ≤ (uD ) , it follows that ar = ap = 0. But then br ∨ ar = br = v = cr = cr ∨ ar , while bp ∨ ap ≺ cp ∨ ap , contradicting that b ≤ c . Thus b ≺ c. We conclude that L is isoform. This completes the proof of Theorem 14.1.

14.6.

Discussion

Regular lattices Sectionally complemented lattices are regular, so we already have a congruence-preserving extension theorem for regular lattices in Chapter 12. We would like to point out that Theorem 14.1 contains this result. Lemma 14.12. Every ﬁnite isoform lattice is regular. Proof. Let L be an isoform lattice. Let Θ and Φ congruences sharing the block A. Then A is also a block of Θ ∧ Φ; in other words, we can assume that Θ ≤ Φ. Let B be an arbitrary Θ block. Since Θ ≤ Φ, there is a (unique) Φ block B containing B. Since L is isoform, A ∼ = B; by ﬁniteness, = B and A ∼ B = B, that is, Θ = Φ. Corollary 14.13. The lattice L of Theorem 14.1 is regular. Permutable congruences Since sectionally complemented lattices have permutable congruences, the congruence-preserving extension result for congruences permutable lattices follows from the result in Chapter 12. We would like to point out that Theorem 14.1 contains also this result. This statement follows from Lemma 14.14. The lattice L of Theorem 14.1 is congruence permutable.

174

14. Isoform Lattices

Proof. This is obvious since the congruences of L are congruences of S, which is a direct product of simple lattices. Problem 2 of Gr¨ atzer, Quackenbush, and Schmidt [53] raised the question whether this is typical. Kaarli [95] provided an aﬃrmative solution: Theorem 14.15. Every ﬁnite uniform lattice is congruence permutable. Kaarli also described a countably inﬁnite isoform lattice that is not congruence permutable. Deterministic isoform lattices Let L be an isoform lattice, let Φ be a congruence of L, and let AΦ be a congruence class of Φ. Then any congruence class of Φ is isomorphic to AΦ . However, if Φ and Ψ are congruences of L, it may happen that AΦ and AΨ as lattices are isomorphic. Let us call an isoform lattice L deterministic if this cannot happen; in other words if Φ = Ψ are congruences of L, then AΦ and AΨ are not isomorphic. Lemma 14.16. The lattice L of Theorem 14.1 can be constructed to be deterministic. Proof. The size |AΦ | is the product of the |SΦ |, Φ ∈ ConM K. Since we can easily construct the SΦ -s so that all |SΦ |-s are distinct primes, the statement follows. Naturally isoform lattices Let L be a ﬁnite lattice. Let us call a congruence relation Θ of L naturally isoform if any two congruence classes of Θ are naturally isomorphic (as lattices) in the following sense: if a ∈ L is the smallest element of the class a/Θ, then x → x ∨ a is an isomorphism between 0/Θ and a/Θ. Let us call the lattice L naturally isoform if all congruences of L are naturally isoform. The lattice L of Theorem 14.1 is not naturally isoform. There is a good reason for it. Theorem 14.17. Let L be a ﬁnite lattice. If L is naturally isoform, then Con L is boolean. Proof. If L is simple, the statement is trivial. Let Θ be a nontrivial congruence relation of L. Let a be the largest element of the Θ-class 0/Θ and let b be the smallest element of the Θ-class 1/Θ. Then obviously a ∨ b = 1. If a ∧ b > 0, then b ∨ 0 = b ∨ (a ∧ b) (= b); therefore, x → x ∨ b is not an isomorphism between 0/Θ and b/Θ. Thus a ∧ b = 0. We prove that L ∼ = id(a) × id(b). For c ∈ L, we get c ∧ b ≡ c ∧ 1 = c (Θ), and so c∧b ∈ c/Θ. If d < c∧b is the smallest element of c/Θ, then the natural

14.6. Discussion

175

isomorphism between 0/Θ and d/Θ would force that c ∧ b = d ∨ x, for some 0 < x ≤ a, contradicting that (c∧b)∧a = 0. The natural isomorphism between 0/Θ and c/Θ yields that c = (c ∧ b) ∨ x, for some unique x ≤ a. Since x ≤ c, clearly, x ≤ c ∧ a. Therefore, (c ∧ b) ∨ (c ∧ a) ≤ c = (c ∧ b) ∨ x ≤ (c ∧ b) ∨ (c ∧ a), ∼ id(a) × id(b). that is, c = (c ∧ b) ∨ (c ∧ a). This proves that L = Thus Θ is the kernel of the projection of L onto id(a) and so has a complement, the kernel of the projection of L onto id(b). We conclude that Con L is boolean. A generalized construction We deﬁned an element v of a ﬁnite lattice A to be a separator if 0 ≺ v ≺ 1. Instead, we can have the more general deﬁnition: a separator v is a doubly irreducible element. Then there is a unique v∗ ≺ v and a unique v ∗ v, and these take over the role of 0 and 1 in the proofs. Let us restate Theorem 14.4: Theorem 14.18. L is a lattice. Let a = ap p∈P , b = bp p∈P ∈ L. Then ⎧ ∗ ⎪ if ap ∨ bp = v and, for some p ≥ p, ⎪ ⎪v , ⎪ ⎪ ⎪ (1∨ ) p is an {a, b}-fork, or ⎨ (a ∨ b)p = (2∨ ) bp ≤ ap and bp ap , or ⎪ ⎪ ⎪ (3∨ ) ap ≤ bp and ap bp ; ⎪ ⎪ ⎪ ⎩a ∨ b , otherwise; p

and

(a ∧ b)p =

p

⎧ ⎪ ⎪ ⎪v∗ , ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩a ∧ b , p p

if ap ∧ bp = v and for some p ≥ p, (1∧ ) p is an {a, b}-fork, or (2∧ ) bp ≥ ap and bp ap , or (3∧ ) ap ≥ bp and ap bp ; otherwise.

Problems Problem 14.1. Develop a theory of (deterministic) isoform lattices. Problem 14.2. Let f (n) be the smallest integer with the property that every lattice K of n elements has a congruence-preserving extension into an isoform lattice of size f (n). Estimate f . Let L be a ﬁnite lattice. A congruence Θ of L is algebraically isoform if, for every a ∈ L, there is a unary algebraic function p(x) that is an isomorphism between 0/Θ and a/Θ. The lattice L is algebraically isoform if all congruences are algebraically isoform. Problem 14.3. Does every ﬁnite lattice have a congruence-preserving extension to an algebraically isoform ﬁnite lattice?

176

14. Isoform Lattices

Problem 14.4. Can we carry out the construction for Theorem 14.1 in case Con K is ﬁnite (as opposed to K being ﬁnite)? Problem 14.5. Is there an analogue of Theorem 14.1 for inﬁnite lattices? By Lemma 14.16, every ﬁnite lattice has a congruence-preserving extension to a deterministic lattice. Can this result be extended to inﬁnite lattices? Problem 14.6. Does every lattice have a congruence-preserving extension to a deterministic lattice?

15 Independence Theorems

15.1.

Results

The following problem was ﬁrst raised in the ﬁrst edition of [26] (Problem II.18): Problem. Let K be a nontrivial lattice and let G be a group. Does there exist a lattice L such that the congruence lattice of L is isomorphic to the congruence lattice of K and the automorphism group of L is isomorphic to G? If K and G are ﬁnite, can L chosen to be ﬁnite? This problem was solved for ﬁnite lattices by Baranski˘ı [3], [4] and Urquhart [120]. We now state the Baranski˘ı–Urquhart theorem: Theorem 15.1 (Independence Theorem). Let D be a nontrivial ﬁnite distributive lattice and let G be a ﬁnite group. Then there exists a ﬁnite lattice L such that the congruence lattice of L is isomorphic to D and the automorphism group of L is isomorphic to G. This is a representation theorem; both published proofs rely heavily on the Dilworth Theorem (Theorem 7.1) and on the representation theorem of ﬁnite groups as automorphism groups of ﬁnite lattices (Birkhoﬀ [7] and Frucht [20] 15.2.5). The topic of this chapter is the congruence-preserving extension variant of Theorem 15.1, which we published in Gr¨ atzer and Schmidt [66]: Theorem 15.2 (Strong Independence Theorem). Let K be a nontrivial ﬁnite lattice and let G be a ﬁnite group. Then K has a congruence-preserving extension L whose automorphism group is isomorphic to G.

178

15. Independence Theorems

There is a stronger form of this theorem; to state it, we need the analogue of the congruence-preserving extension concept for automorphisms. Let K be a lattice. L is an automorphism-preserving extension of K if L is an extension and every automorphism of K has exactly one extension to L; moreover, every automorphism of L is the extension of an automorphism of K. Of course, then the automorphism group of K is isomorphic to the automorphism group of L. Now we state the stronger version of Theorem 15.2: Theorem 15.3 (Strong Independence Theorem, Full Version). Let KC and KA be nontrivial ﬁnite lattices. Let us assume that KC ∩KA = {0}. Then there exists a lattice L such that the following conditions hold: (1) L is a ﬁnite, atomistic, {0}-extension of both KA and KC . (2) L is a congruence-preserving extension of KC . (3) L is an automorphism-preserving extension of KA . Of course, then the congruence lattice of L is isomorphic to the congruence lattice of KC , and the automorphism group of L is isomorphic to the automorphism group of KA . Notice how the topic is shifting. In Chapters 12 and 13, we seek congruence-preserving extensions into ﬁrst-order deﬁnable classes: sectionally complemented and semimodular lattices. In Chapter 14, we deal with the properties of congruences: uniform and isoform congruences. In this chapter, for the ﬁrst time, we go “outside” the lattice, looking at automorphism groups. Naturally, the techniques that worked so well no longer apply, in particular, we do not use cubic extensions. Chopped lattices, however, continue to play a crucial role.

15.2.

Proof-by-Picture

15.2.1

Frucht lattices

By a graph C = V, E, we mean a nonempty set V of vertices and a set E of edges; every e ∈ E is a two-element subset of V . Figure 15.1 illustrates how we draw graphs: the vertices are represented by small circles ; the circles representing two distinct vertices x and y are connected by a line segment iff {x, y} is an edge. An automorphism α of C is a bijection of V preserving the edges, that is, {x, y} is an edge iff {xα, yα} is an edge, for x = y ∈ V . The automorphisms of C form a group, denoted by Aut C. Now let C be a graph V, E, and—as in R. Frucht’s [19]—we order the set Frucht C = {0, 1} ∪ V ∪ E by 0 < v, e < 1, for all v ∈ V and e ∈ E, and v < e, for v ∈ V and e ∈ E. iff v ∈ e. The order Frucht C is a lattice; it is in fact atomistic and of length 3. See Figure 15.1 for a small example. It is clear

15.2. Proof-by-Picture

179

1 c w y

d

C

x

u

v

a

b

x

y

w

v b

c

d

u

Frucht C 0

a

Figure 15.1: Constructing the Frucht lattice. from the picture that C is coded into Frucht C, so Aut(Frucht C) = Aut C. The lattice Frucht C is almost always simple (see Section 15.4). So we obtain the following result (Frucht [20]): Theorem 15.4. For every graph C, there is an atomistic lattice F of length 3 with Aut C ∼ = Aut F . 15.2.2

An automorphism-preserving simple extension

A graph C = V, E is rigid iff it only has the identity map as an automorphism. Rigid graphs are easy to construct. Figure 15.2 shows a rigid graph with 7 vertices, a 6-cycle with a chord and a tail. Replace 6 with any n ≥ 6, and one gets a rigid graph Rn , deﬁned on the set {r0n , r1n , . . . , rnn }, with n + 1 elements; the edges are {r0n , r1n } (tail), {r1n , r2n }, {r3n , r2n }, . . . , {rnn , r1n } (cycle), and {r1n , r3n } (chord). These graphs have the important property that if n = m, then Rn has no embedding into Rm .

r46 r36

r56

r26

r66 r16 r06

Figure 15.2: The rigid graph, R6 .

180

15. Independence Theorems

We start the Proof-by-Picture of Theorem 15.3 with the following modest step: Theorem 15.5. The lattice KA has a ﬁnite, atomistic, simple, automorphism- and {0}-preserving extension LA with an atom u that is ﬁxed under all the automorphisms.

1

1

LA

KA F1 a

b

u a

b Fa

Fb

Figure 15.3: The lattice LA for KA To illustrate the construction, let us take KA = B2 + C1 be the fourelement boolean lattice with atoms a and b and with a new unit element, 1; see Figure 15.3. We take three copies: Fa , Fb , and F1 of the Frucht lattice Frucht R6 and replace the interval [0, a] of KA with Fa , the interval [0, b] of KA with Fb , and adjoin F1 by identifying the zero of F1 with the zero of KA , the unit of F1 with the unit of KA , as depicted in Figure 15.3. In general, for any join-irreducible element x, we attach a copy Fx of F so that the zero of Fx is identiﬁed with the zero of KA and the unit of Fx is identiﬁed with x; all the other elements of Fx are incomparable with the other elements of [0, x]. Clearly, LA is a lattice; it is atomistic because F is; it is simple because of all the complements. Since it is atomistic, automorphisms are determined by how they act on atoms. It now follows that LA is an automorphism-preserving extension, because F is rigid. 15.2.3

A congruence-preserving rigid extension

The next step constructs an extension LC of the lattice KC (of Theorem 15.3) that retains the congruences but destroys the automorphisms. Theorem 15.6. The lattice KC has a ﬁnite, atomistic, rigid, congruencepreserving, {0}-extension LC . By Theorem 12.1, we can take a sectionally complemented, congruence of the lattice KC , with zero, 0, and unit, 1C . preserving, {0}-extension KC . With every atom ai in KC , we Let {a1 , . . . , an } be the set of atoms of KC associate the (rigid Frucht) lattice Fi = Frucht Ri+w , where w = |KC |, with zero, 0i , and unit, 1i ; let pi = r0i .

15.2. Proof-by-Picture

1C

181

12

11 KC

M2

p1 a1 F1

a2

p2 F2

0 Figure 15.4: The second step in the construction of the chopped lattice M . Let M1 be the chopped lattice we obtain merging KC and the lattice F1 by identifying the zero of KC with the zero of F1 and by identifying the atom a1 of KC with the atom p1 of F1 ; so KC ∩ F1 = {0, a1 = p1 }, in M1 , as illustrated in Figure 15.4. Now we merge F2 and M1 , with the atoms a2 and p2 , the same way, to obtain the chopped lattice M2 ; see Figure 15.4. In n steps, we obtain the chopped lattice M = Mn . The lattice LC is deﬁned as Id M .

15.2.4

Merging the two extensions

We are given the lattices LA , constructed in Section 15.2.2, and LC , constructed in Section 15.2.3. We assume that LA ∩ LC = {0}. We further assume that LA LC , otherwise, we can take L = LA . Let v be any atom of LC . Let u denote the atom of LA that is ﬁxed by all automorphisms. We construct the chopped lattice N by merging LA and LC , by identifying the zeros of LA and LC , and identifying u with v; see Figure 15.5.

N

LA u v

LC

0 Figure 15.5: Merging the two lattices, LA and LC .

182

15. Independence Theorems

g2 x

y

3 4 vx,y 4 x

2 1

ux,y

y

Figure 15.6: Substituting a labeled edge with a graph. Under some very mild conditions, the lattice L = Id N satisﬁes the conditions of the Strong Independence Theorem, Full Version. 15.2.5

The representation theorems

To prove the Strong Independence Theorem, we need a representation theorem for groups: Theorem 15.7. Every group G is isomorphic to the automorphism group of a lattice KA . We use the term digraph for a directed graph (the edges have a direction) without multiple edges—for each pair of vertices v, w, there is at most one edge from v to w. The digraphs we consider here will also have no loops, that is, the ends of any edge will be distinct. Labeled digraph means that we have a set X of labels and a surjective mapping λ from the directed edges to X. Step 1: The labeled Cayley digraph, Cayley G. We start with the four-element commutative group G = {1, a, b, c}, with multiplication: a2 = c2 = b, b2 = 1, ab = c, bc = a, and ac = 1. Figure 15.7 shows the associated labeled Cayley digraph, Cayley G, with the set of labels X = G − {1}. We can carry out this construction for any ﬁnite group G. The graph Cayley G is deﬁned on the set G, and x = y are connected with an arrow from x to y labeled by z (x, y ∈ G, z ∈ G − {1}) iff y = xz. So the labeled digraph represents right multiplication, hence, the automorphisms of this structure are the left multiplications: λu : x → ux (u, x ∈ G). The map u → λu is an isomorphism between G and the automorphism group Aut(Cayley G) of the labeled digraph. Step 2: The Cayley graph, Cay G. Let G − {1} = {g1 , . . . , gn }. For 1 ≤ m ≤ n, we denote by m the graph deﬁned on the set {1, . . . , m + 2}, where i and j are joined by an edge, for

15.3. Formal proofs

183

1 ≤ i, j ≤ m + 2 iff i = j. So m is the full graph on m + 2 elements without loops or double edges. In Cayley G, we replace the labeled directed edges by (undirected, not labeled) graphs (without loops) as follows: Let the directed edge from x to y labeled by gm (x, y ∈ G, 1 ≤ m ≤ n) be replaced by the graph deﬁned on {x, y, ux,y , vx,y } ∪ m, where m is a full subgraph and there are edges between x and 1 and between y and 1, and between ux,y and vx,y , as illustrated by Figure 15.6; let S(x, y) denote this graph. Let Cay G denote the graph we obtain by so replacing all labeled edges. c

1

a a b

c

a

a

c

b a c

b c

Figure 15.7: A labeled Cayley digraph. Let α be an automorphism of Cay G. It is clear that a vertex which is the endpoint of exactly one edge is of the form vx,y , for x = y ∈ G, so α permutes the vx,y -s. Since vx,y determines Sx,y , the map α permutes the Sx,y -s. Which means that α uniquely deﬁnes an automorphism of Cayley G, and conversely. Now by Theorem 15.4, the lattice KA = Frucht(Cay G) will do for Theorem 15.7. The presentation in this section follows Gr¨ atzer and Lakser [40].

15.3.

Formal proofs

In this section we go through the constructions and theorems of Section 15.2 and provide the details, thereby proving the three theorems of Section 15.1. 15.3.1

An automorphism-preserving simple extension

In this section we provide a proof of Theorem 15.5. Let F = Frucht R6 , which is a ﬁnite, simple, rigid, atomistic lattice of length 3. For every join-irreducible a ∈ KA , we take a copy Fa of the lattice F with zero 0a and unit 1a . We form the disjoint union ¯ = KA ∪ ( Fa − {0a , 1a } | a ∈ KA − {0} ) L

184

15. Independence Theorems

and we identify 0a with 0 and 1a with a. ¯ we deﬁne x ∧ y and x ∨ y as follows: For x, y ∈ L, ¯ (i) Let K and all the Fa -s be sublattices of L. (ii) If x ∈ Fa − K, y ∈ Fb − K, a = b, then x ∧ y = 0 and x ∨ y = a ∨ b. (iii) If x ∈ Fa − K, y ∈ K − Fa , then x, if a ≤ y, x∧y = 0, otherwise; and x ∨ y = a ∨ y. And symmetrically, for x ∈ K − Fa and y ∈ Fa − K. ¯ is a lattice containing K and all It is an easy computation to show that L the Fa , a ∈ J(K), as sublattices. ¯ to obtain LA ; u is a common comFinally, we adjoin an element u to L plement to all the elements of L − {0, 1}. We prove that LA satisﬁes the conditions of Theorem 15.5. The lattice LA is atomistic. Every Fa is atomistic, therefore, so is LA . The lattice LA is simple. Indeed, since LA is atomistic, every congruence relation is of the form con(0, a). So it is enough to prove that con(0, a) is the unit congruence ι, for all 0 < a. If a = u, then a ≡ 0 (Θ) implies that u ≡ 1 (Θ). Let b ∈ LA − {0, u, 1}; then from the last congruence we obtain that b ≡ 0 (Θ). Since we can write 1 as the join of two such elements b, we conclude that 1 ≡ 0 (Θ), and so Θ = ι. A similar computation shows that u ≡ 0 (Θ) implies that Θ = ι, so LA is simple. Now we compute that LA is a automorphism-preserving extension of K. First, let α be an automorphism of K. We deﬁne the extension α ˆ of α to LA . If a, b ∈ K and aα=b, then let α ˆ map Fa onto Fb under the unique α = u. Then α ˆ is obviously isomorphism between Fa and Fb . Finally, deﬁne uˆ an extension of α to LA . Second, we prove that every automorphism β of LA is of the form α ˆ for a suitable automorphism α of K. Let a ∈ J(K). We are going to deﬁne aα. Let p be an atom of Fa , for some a ∈ J(K); then p is an atom of LA not covered by 1. Therefore, pβ is an atom of LA not covered by 1. Now observe that every atom of LA not covered by 1 is an atom of some Fb , with b ∈ J(K). Therefore, pβ ∈ Fb for some b ∈ J(K). Let q be an atom of Fa such that {p, q} is an edge in the graph V, E. Then p ∨ q ∈ Fa covers p and q. Therefore, (p ∨ q)β covers pβ and qβ. Fb is of length 3, and an atom of Fb is covered only by elements of Fb by the construction of LA . Since pβ is an atom in Fb , it follows that (p ∨ q)β ∈ Fb , and so qβ ∈ Fb . The graph deﬁning F is connected, so we conclude that the image of any atom of Fa is in Fb . The lattice Fa is atomistic, so we obtain that Fa β is a sublattice of Fb . Since Fa and Fb are copies of the same ﬁnite rigid lattice F , it follows that β maps Fa onto Fb ; in particular, a to b. We deﬁne aα = b.

15.3. Formal proofs

185

This deﬁnes α on J(K); adding 0α = 0, we now have α deﬁned on K. It is obvious that α is an automorphism of K and α ˆ = β, by the rigidity of F . 15.3.2

A congruence-preserving rigid extension

We continue with the proof of Theorem 15.6 by verifying the properties of the lattice LC = Id M . The lattice LC is obviously a ﬁnite, atomistic, {0}-extension of KC . To show that LC is rigid, let α be an automorphism of LC . We regard M ⊆ LC . Observe that M and LC have the same set of atoms. Observation: Let x = y be atoms. If x ∈ KC − Fi and y ∈ Fi − KC , or if x ∈ Fj − Fi and y ∈ Fi − Fj , where 1 ≤ i = j ≤ n, then the interval [0, x ∨ y] of LC is a four-element boolean sublattice of LC . Now assume that there is an i with 1 ≤ i ≤ n and an atom rji ∈ Fi , with 1 ≤ j ≤ i + w such that rji α ∈ / Fi . We distinguish two cases. Case 1: rji α ∈ KC . Consider the atoms rki with |j − k| > 1. Then rji ∨ rki = 1i , so the interval [0, rji ∨ rki ] is of length 3. Therefore, the interval [0α, (rji ∨ rki )α] = [0, rji α ∨ rki α] is also of length 3, so by the Observation above, rki α ∈ KC . So α maps all these atoms into KC , and therefore, α is an embedding of Fi into KC . This contradicts that |Fi | > |KC |.

Case 2: rji α ∈ Fk , with k = i. Arguing as in Case 1, we get that α is an embedding of Fi into Fk . This contradicts that the graph Ri+w has no embedding into the graph Rk+w . The completes the veriﬁcation that LC is rigid. Finally, the lattice LC is a congruence-preserving extension of KC . Indeed, the chopped lattice M is a congruence-preserving extension of KC by a trivial induction on n, using Corollary 4.7, and LC is a congruence-preserving extension of KC by Theorem 4.6. This completes the proof of Theorem 15.6. 15.3.3

Proof of the independence theorems

We are given the two ﬁnite lattices KC and KA with KC ∩ KA = {0}. Let LC be the lattice provided for KC by Sections 15.2.3 and 15.3.2; the unit of LC is denoted by 1C . Let LA be the lattice provided for KA by Sections 15.2.2 and 15.3.1; the unit of LA is denoted by 1A . We can assume that LA ∩ LC = {0}. We can further assume that LA LC . Indeed, if LA ∼ = LC , then we can take L = LA . We construct the chopped lattice N and the lattice L = Id N as in Section 15.2.4; see Figure 15.5. Since LA is simple, by Corollary 4.7, we get that N is a congruence-preserving extension of LC , and LC is a congruence-preserving extension of N by Theorem 4.6, so L is a congruence-preserving extension of LC . To verify the automorphism properties of L, we need two lemmas:

186

15. Independence Theorems

Lemma 15.8. Let U be a chopped lattice with two maximal elements m1 and m2 . If p = m1 ∧ m2 is an atom, then Id U is a union of four parts: U1 = id(m1 ), U2 = id(m2 ), U1,2 = (id(m1 ) − [p, m1 ]) × (id(m2 ) − [p, m2 ]), U 1,2 = [p, m1 ] × [p, m2 ]. Proof. This is easy: if j1 , j2 is a compatible vector (see Section 4.2), then either j1 ∧ p = j2 ∧ p = 0 or j1 ∧ p = j2 ∧ p = p. In the ﬁrst case, j1 , j2 ∈ U1 if j2 = 0, j1 , j2 ∈ U2 if j1 = 0, and j1 , j2 ∈ U1,2 if j1 = 0 and j2 = 0. In the second case, j1 ≥ p1 and j2 ≥ p2 , so j1 , j2 ∈ U 1,2 . Lemma 15.9. Using the assumptions and notation of Lemma 15.8, let αi be an automorphism of Ui , for i = 1, 2. If p = pα1 = pα2 , then there is a unique automorphism α ˆ of U that extends both α1 and α2 . Proof. Let p = pα1 = pα2 . Then deﬁne α ˆ on a compatible vector j1 , j2 by j1 , j2 ˆ α = j1 α1 , j2 α2 . It is easy to see that α ˆ is an automorphism of U . Conversely, let α ˆ be an automorphism of U extending both α1 and α2 . The compatible pair j1 , j2 is either in U 1,2 , in which case it satisﬁes j1 , j2 = j1 , p ∨ p, j2 or j1 , j2 ∈ / U 1,2 , in which case it satisﬁes j1 , j2 = j1 , 0 ∨ 0, j2 . Either one of the equations implies that j1 , j2 ˆ α = j1 α1 , j2 α2 . Let us assume now that the Frucht lattice F used in the construction of LA satisﬁes the condition |F | > |LA |, |LC |. To prove that L is an automorphismpreserving extension, we have to prove two claims: Claim 1. Every automorphism of KA has exactly one extension to L. Proof. Let α1 be an automorphism of KA ; let α2 be the identity map on LC . The atom p of KA ∩KC is the atom u of KA ﬁxed by all automorphisms of KA , so p = pα1 = pα2 holds. By Lemma 15.9, α1 (and α2 ) has an extension to an automorphism α ˆ 1 of L. To get the uniqueness, let α1 extend α1 to L. Then pα1 = p. Since α1 permutes the atoms of LA and keeps p ﬁxed, it follows that it permutes the atoms of L other than the atoms of LA , that is, it permutes the atoms of LC . So α1 restricted to LC is an automorphism of LC . Since LC is rigid, α1 restricted to LC is the identity map. So α1 = α ˆ 1 of Lemma 15.9, and it is unique. Claim 2. Every automorphism of L is the extension of an automorphism of KA . Proof. Let α be an automorphism of L. Let us take the set H of all element h of L satisfying the condition that in L, the interval [0, h] contains a copy of F such that 0 is the zero of F and h is the unit of F . We claim that H = J(KA ).

15.4. Discussion

187

Indeed, If h ∈ J(KA ), then the interval [0, h] contains a copy of F by the construction of LA ; see Sections 15.2.2 and 15.3.1. If h ∈ L, then h is in one of the regions described in Lemma 15.8, namely, U1 = LA , U2 = LC , U1,2 , or U 1,2 . Let h ∈ H but h ∈ / J(KA ). We distinguish four cases. Case 1: h ∈ U1 = LA . Then h ∈ LA so either h ∈ KA or h ∈ Fa , for some a ∈ J(KA ) with h < a. If h ∈ KA , then h is join-reducible, so the copy of F such that 0 is the zero of F and h is the unit of F , must be in KA , contradicting the assumption: |KA | < |F |. If h ∈ Fa , for some a ∈ J(KA ) with h < a, then the copy of F must be disjoint to Fa ; therefore, F ⊆ KA , the same contradiction as in the previous case. Case 2: h ∈ U2 = LC . Then this contradicts the assumption: |KC | < |F |. Case 3: h ∈ U1,2 . Then [0, h] in L is a direct product, contradicting that F is directly irreducible. Case 4: h ∈ U 1,2 , but h ∈ / U1 and h ∈ / U2 . Then v = 1A , h2 , for some h2 ∈ [v, 1V ], with h2 = u, since u is a dual atom of LA . So the dual atoms of F in [0, h] with at most one exception come from the dual atoms of [0, h2 ] in LC , yielding again a contradiction with |F | > |LC |. This concludes the proof of H = J(KA ). Since α must permute H, we conclude that α permutes J(KA ), and so α restricted to KA is an automorphism of KA . By the construction of LA we immediately get that α restricted to LA is an automorphism α1 of LA . These claims show that lattice L = Id N satisﬁes the conditions of the Strong Independence Theorem, Full Version.

15.4.

Discussion

Automorphism groups It was ﬁrst proved in Birkhoﬀ [7] that every ﬁnite group is isomorphic to the automorphism group of a ﬁnite (distributive) lattice. R. Frucht in [19] constructed a length 3 lattice with a given automorphism group, as described in Section 15.2.1. For Frucht’s construction, we need a graph with a given automorphism group. Such a graph was constructed in [19]; see also Sabidussi [104]. For a general introduction to this topic, see the monograph Pultr and Trnkov´ a [102].

188

15. Independence Theorems

Two groups In the style of Part V, in [40], Gr¨ atzer and Lakser proved a “two group” theorem: Theorem 15.10. Let G, G be groups and let η : G → G be a group homomorphism. Then there is a simple lattice H of length at most 6 with an ideal H = id(iH ) of length at most 3 which is also simple. There are isomorphisms τH : G → Aut H and

τH : G → Aut H

such that, for each g ∈ G, the automorphism gτH of H restricts to the auto morphism gητH of H . Furthermore, if G and G are ﬁnite, then so is the lattice H. Simple Frucht Lattices In [40], G. Gr¨ atzer and H. Lakser observe that the Frucht lattices are almost always simple. The Frucht lattice F constructed from the graph C = V, E is simple iff the following condition holds: For v ∈ V , there are a1 , a2 ∈ E with v ∈ / a1 , a2 and a1 ∩ a2 = ∅. Problems We do not have an independence theorem for lattices with any special properties. Let us mention two possibilities: Problem 15.1. Is there an independence theorem for sectionally complemented lattices? Problem 15.2. Is there an independence theorem for semimodular lattices? The combinatorial questions are all open: Problem 15.3. What is the size of the lattice L we construct for Theorem 15.2? What is the minimum size of a lattice L—as a function of |K| and |G|—satisfying Theorem 15.2? Problem 15.4. What is the size of the lattice L we construct for Theorem 15.3? What is the minimum size of a lattice L—as a function of |KC | and |KA |—satisfying Theorem 15.3? Theorem 15.1 can be proved also for general lattices, see Gr¨ atzer and Wehrung [85]. Unfortunately, this proof is very long and complex. Problem 15.5. Is there a shorter proof for the Independence Theorem for general lattices?

16 Magic Wands

16.1.

Constructing congruence lattices

16.1.1

Bijective maps

A typical way of constructing an algebra B with a given congruence lattice C is to construct an algebra A with a much larger congruence lattice and then “collapsing” suﬃciently many pairs of congruences of the form con(a, b) and con(c, d) in B, so that the congruence lattice “shrinks” to C. To do this we need a “magic wand” that will make a ≡ b equivalent to c ≡ d. Such a magic wand may be a pair of partial operations f and g such that f (a) = c, f (b) = d, and g(c) = a, g(d) = b. This is the start of the Congruence Lattice Characterization Theorem of Universal Algebras of Gr¨ atzer and Schmidt (see [58], and also [22] and [25].) If we want to construct a lattice L with a given congruence lattice C, how do we turn the action of the “magic wand” into lattice operations? To construct a simple modular lattice, Schmidt [106] started with the rational interval K = [0, 1] and by a “magic wand” he required that all [a, b] (0 ≤ a < b ≤ 1) satisfy that a ≡ b be equivalent to 0 ≡ 1. The action of the magic wand was realized with the gadget M3 [K] (see Section 5.3). In Sections 16.2 and 16.3 we prove that one can apply the magic wand to arbitrary lattices with zero, see Gr¨ atzer and Schmidt [75]. If we are considering a “magic wand” (an extension L of the lattice K) that will realize that a ≡ b be equivalent to c ≡ d in the lattice L, we immediately notice that we have to say something about the intervals [a, b] and [c, d]. For instance, if a ≡ b (Θ ∨ Ψ) in K, for congruences Θ and Φ of L, then c ≡ d (Θ ∨ Ψ), for congruences Θ = conL (Θ) and Φ = conL (Φ) of L, therefore, the

190

16. Magic Wands

sequence in [a, b] that forces a ≡ b (Θ ∨ Ψ) in K (see Theorem 1.2) must somehow be mapped in L to a sequence in [c, d]L to force c ≡ d (Θ ∨ Ψ). So for lattices, “magic wands” must act on intervals, not on pairs of elements. To set up “magic wands”—as (convex) extensions—for lattices formally, let K be a bounded lattice, let [a, b] and [c, d] be intervals of K, and let ϕ : [a, b] → [c, d] be an isomorphism between these two intervals. We can consider ϕ and ϕ−1 as partial unary operations. Let us call a congruence Θ of K a ϕ+ congruence iff Θ satisﬁes the Substitution Property with respect to the partial unary operations ϕ and ϕ−1 . Let Kϕ+ denote the partial algebra obtained from K by adding the partial operations ϕ and ϕ−1 . Thus a congruence relation of the partial algebra Kϕ+ is the same as a ϕ+ -congruence of the lattice K. We call an extension L of the lattice K a ϕ+ -congruence-preserving extension of K if a congruence of K extends to L iff it is a ϕ+ -congruence and every ϕ+ -congruence of K has exactly one extension to L. As a special case, we get the well-known concept of a congruence-preserving extension (in case, ϕ is trivial). Let us call ϕ (resp., ϕ−1 ) algebraic iff there is a unary algebraic function p(x) such that xϕ = p(x), for all x ∈ [a, b] (resp., xϕ−1 = p(x), for all x ∈ [c, d]). In Gr¨ atzer and Schmidt [75], we prove the following result: Theorem 16.1. Let K be a bounded lattice, let [a, b] and [c, d] be intervals of K, and let ϕ : [a, b] → [c, d] be an isomorphism between these two intervals. Then K has a ϕ+ -congruence-preserving convex extension into a bounded lattice L such that both ϕ and ϕ−1 are algebraic in L. In particular, the congruence lattice of the partial algebra Kϕ+ is isomorphic to the congruence lattice of the bounded lattice L. So the lattice L constructed in this result is the magic wand for ϕ. 16.1.2

Surjective maps

Gr¨ atzer, Greenberg, and Schmidt [32] generalized to surjective maps the approach of Section 16.1.1. Let K be a bounded lattice, let [a, b] and [c, d] be intervals of K, let ϕ be a homomorphism of [a, b] onto [c, d]. We can consider ϕ as a partial unary operation. Let us call a congruence Θ of K a ϕ-congruence iff Θ satisﬁes the Substitution Property with respect to the partial unary operation ϕ, that is, x ≡ y (Θ) implies that xϕ ≡ yϕ (Θ), for all x, y ∈ [a, b]. Let Kϕ denote the partial algebra obtained from K by adding the partial operation ϕ. Thus a congruence relation of the partial algebra Kϕ is the same as a ϕ-congruence of the lattice K. We call an extension L of the lattice K a ϕ-congruence-preserving extension of K if a congruence of K extends to L iff it is a ϕ-congruence and every

16.2. Proof-by-Picture for bijective maps

191

ϕ-congruence of K has exactly one extension to L. If L is a ϕ-congruencepreserving extension of K, then the congruence lattice of the partial algebra Kϕ is isomorphic to the congruence lattice of the lattice L. Theorem 16.1 generalizes as follows: Theorem 16.2. Let K be a bounded lattice, let [a, b] and [c, d] be intervals of K, and let ϕ be a homomorphism of [a, b] onto [c, d]. Then K has a ϕcongruence-preserving convex extension into a bounded lattice L such that ϕ is algebraic in L. In particular, the congruence lattice of the partial algebra Kϕ is isomorphic to the congruence lattice of the bounded lattice L. If K is ﬁnite, then L can be constructed as a ﬁnite lattice. So L realizes the “magic wand”: a ≡ b implies that c ≡ d in L. To illustrate the use of such a result, we reprove the Dilworth Theorem (Theorem 7.1). Let D be a ﬁnite distributive lattice and P = J(D). Let K be the boolean lattice with atoms ai , i ∈ P . For i > j in P , deﬁne ϕi,j as the only homomorphism of [0, ai ] onto [0, aj ]. Apply Theorem 16.2 to these homomorphisms one at a time to construct a lattice L whose congruence lattice is isomorphic to D.

16.2.

Proof-by-Picture for bijective maps

In this section we present the Proof-by-Picture of Theorem 16.1; we build the lattice L. We use two gadgets: the boolean triple construction of Chapter 5, and its special intervals of Section 5.4. Let the bounded lattice K, the intervals [a, b] and [c, d], and let the isomorphism ϕ : [a, b] → [c, d] be given as in Theorem 16.1. We start the construction with four gadgets, which are assumed to be pairwise disjoint, see the top left of Figure 16.1: The four gadgets. (i) A = M3 [K]. Let {0A , p1 , p2 , p3 , 1A } be the spanning M3 in A—as in Lemma 5.1.(iv). (ii) B = M3 [a, b], with the spanning M3 : {0B , q1 , q2 , q3 , 1B }. (iii) The lattice Ka,b —as in Lemma 5.19. (iv) The lattice Kc,d —as in Lemma 5.19. Some notation: An element of one of the four gadgets is described as a triple x, y, z ∈ K 3 belonging to the particular gadget. Note that a triple x, y, z may belong to two or more gadgets. If it is not clear the element of

192

16. Magic Wands

A = M3 [K]

A = M3 [K] p2

p1

p1

p3 ϕ2

Ka,b

Fc,d

Fa,b Ic,d

Kc, d

D

q1

U q2

p3

I ψ

Ic,d

Kc, d

Ia,b

ϕ1 q1

p2 Fc,d

Ka,b

Fa,b Ia,b

V

q2

q3

q3

B = M3 [a, b]

B = M3 [a, b] The four gadgets.

Two gluings.

A = M3 [K]

L p2

p1 Fa,b

Ka,b

p3 Fc,d

Ia,b q1

Ic,d q2

q3

Kc, d

B = M3 [a, b] The ﬁnal gluing.

Figure 16.1: Constructing the lattice L.

which gadget a triple is representing, we will make it clear with subscripts: x, y, zA , for x, y, z as an element of A, x, y, za,b , for x, y, z as an element of Ka,b , x, y, zc,d , for x, y, z as an element of Kc,d , x, y, zB ,

for x, y, z as an element of B.

We start the construction by gluing together B and Ka,b to obtain the lattice U , see the top right of Figure 16.1: Two gluings. In B, we use the

16.2. Proof-by-Picture for bijective maps

193

ﬁlter fil(q1 ) = { 1, x, x | a ≤ x ≤ b }, while in Ka,b we utilize the ideal Ia,b = { 0, x, 0 | a ≤ x ≤ b }, and we consider the isomorphism ϕ1 : 1, x, xB → 0, x, 0a,b ,

x ∈ [a, b],

between the ﬁlter fil(q1 ) of B and the ideal Ia,b of Ka,b to glue B and Ka,b together to obtain the lattice U . Similarly, we glue Kc,d and A over the ﬁlter Fc,d = { x, d, x ∧ d | x ∈ K } of Kc,d and the ideal id(p3 ) = { 0, 0, x | x ∈ K } of A, with respect to the isomorphism ϕ2 : x, d, x ∧ dc,d → 0, 0, xA ,

x ∈ K,

to obtain the lattice V . In U , we deﬁne the ﬁlter F = [q3 , 1B ] ∪ Fa,b , which is the union of [q3 , 1B ] and Fa,b , with the unit of [q3 , 1B ] identiﬁed with the zero of Fa,b . In V , we deﬁne the ideal I = Ic,d ∪ [0A , p1 ], which is the union of Ic,d and [0A , p1 ], with the unit of Ic,d identiﬁed with the zero of [0A , p1 ]. Next we set up an isomorphism ψ : F → I. Since [q3 , 1B ] = { x, x, b | a ≤ x ≤ b } and Ic,d = { 0, x, 0 | c ≤ x ≤ d }, we deﬁne ψ on [q3 , 1B ] by ψ : x, x, bB → 0, xϕ, 0c,d ,

x ∈ [a, b],

194

16. Magic Wands

where ϕ : [a, b] → [c, d] is the isomorphism given in Theorem 16.1. This “twist” by ϕ—hidden in the last displayed formula—is the pivotal idea of the proof. We deﬁne ψ on Fa,b by ψ : x, b, x ∧ ba,b → x, 0, 0A ,

x ∈ K.

It is clear that ψ : F → I is well deﬁned and it is an isomorphism. Finally, we construct the lattice L of Theorem 16.1 by gluing U over I with V over F with respect to the isomorphism ψ : F → I, see the bottom half of Figure 16.1: The ﬁnal gluing. By Lemma 5.1.(iii), the map x → x, 0, 0A is an isomorphism between the lattice K and the ideal id(p1 ) of the lattice A; this gives us a convex embedding of K into A. We identify K with its image, and regard the lattice K as a convex sublattice of the lattice A and therefore of the lattice L. So the lattice L is a convex extension of the lattice K. We have completed the construction of the bounded lattice L of Theorem 16.1. Now we have to show that ϕ and ϕ−1 are algebraic in the lattice L. We deﬁne p(x) = ((((((x∧a, a, ba,b )∨b, a, aa,b )∧q2 )∨d, c, cc,d )∧c, c, dc,d )∨p2 )∧p1 . Figure 16.2 shows that xϕ = p(x), for all x ∈ [a, b]. The proof for ϕ−1 is similar.

16.3.

Veriﬁcation for bijective maps

We now verify that the lattice extension L of K satisﬁes the conditions stated in Theorem 16.1. First, we will describe the congruences of L. We need some notation: For a congruence Θ of K, ΘA denotes the congruence Θ3 restricted to A; Θa,b denotes the congruence Θ3 restricted to Ka,b ; Θc,d denotes the congruence Θ3 restricted to Kc,d ; ΘB denotes the congruence Θ3 restricted to B. Let us start with U . The lattices B and Ka,b are glued together over fil(q1 ) and Ia,b with the isomorphism ϕ1 : 1, x, xB → 0, x, 0a,b , and obviously 1, x, x ≡ 1, y, y (Θ3 ) iff 0, x, 0 ≡ 0, y, 0 (Θ3 ). Hence by Lemma 2.9, r r the congruences of U are of the form ΘB ◦ Θa,b (the symbol ◦ is deﬁned in Section 2.4). The lattice B is a congruence-preserving extension of fil(q1 ) (formed in B) by Theorem 5.3; therefore, by Lemma 3.14 and Corollary 5.18, the lattice U

16.3. Veriﬁcation for bijective maps

195

x6

L p1

p2

d b

p3

x7 x

c x4

x2 b, a, a a,a, b

a x1

x5

d,c, c c, c, d

c, c, c

a,a, a q2

q1

q3

x3

Figure 16.2: The seven steps. is a congruence-preserving extension of Ka,b , which, in turn, is a congruencepreserving extension of Fa,b (∼ = K). So U is a congruence-preserving extension of Fa,b (∼ = K). Similarly, V is a congruence-preserving extension of A, which, in turn, is a congruence-preserving extension of [0A , p1 ] = K, so V is a congruence-preserving extension of [0A , p1 ] = K. We glue U and V together over F and I over ψ; equivalently, we identify the ﬁlter Fa,b = { x, b, x ∧ b | x ∈ K } ⊆ U of U with [0A , p1 ] = { x, 0, 0 | x ∈ K } ⊆ A, and note that for any congruence Θ of K, x, b, x ∧ b ≡ y, b, y ∧ b

(Θ3 )

iff

x, 0, 0 ≡ y, 0, 0 (Θ3 ),

so Θ3 restricted to Fa,b is mapped by ψ to Θ3 restricted to [0A , p1 ]—and we

196

16. Magic Wands

identify the ﬁlter [q3 , 1B ] = { x, x, b | a ≤ x ≤ b } of B with the ideal Ic,d = { 0, x, 0 | c ≤ x ≤ d } of Kc,d by identifying x, x, b with 0, xϕ, 0, for x ∈ [a, b]. So in the lattice L, x, x, 1 ≡ y, y, 1

iff

0, xϕ, 0 ≡ 0, yϕ, 0;

translating this back to K, we obtain that x ≡ y (Θ) iff xϕ ≡ yϕ (Θ). This condition is equivalent to the statement that Θ has the Substitution Property with respect to the partial unary operations ϕ and ϕ−1 . This proves, on the one hand, that if Θ extends to the lattice L, then Θ is a ϕ+ -congruence, and, on the other hand, that a ϕ+ -congruence Θ extends uniquely to L, that is, the lattice L is a ϕ+ -congruence-preserving extension of the lattice K. Second, we have to show that ϕ and ϕ−1 are algebraic in the lattice L. We deﬁne p(x) = ((((((x∧a, a, ba,b )∨b, a, aa,b )∧q2 )∨d, c, cc,d )∧c, c, dc,d )∨p2 )∧p1 . For x ∈ [a, b], we want to compute p(x). There are seven steps in the computation of p(x) (see Figure 16.2): x = x, 0, 0A = x, b, xa,b x1 = x ∧ a, a, b

(in A and in Ka,b ),

x2 = x1 ∨ b, a, a x3 = x2 ∧ q2

(computed in Ka,b ), (computed in U ),

x4 = x3 ∨ d, c, c

(computed in L),

x5 = x4 ∧ c, c, d

(computed in Kc,d ),

x6 = x5 ∨ p2 x7 = x6 ∧ p1

(computed in V ),

(computed in Ka,b ),

(computed in A).

Our goal is to prove that x7 = xϕ. By the deﬁnition of ψ, when gluing U and V together, we identify x = x, 0, 0 ∈ A with x, b, x ∧ b = x, b, x ∈ Ka,b , so x = x, b, x ∈ Ka,b . Therefore, x1 = x, b, x ∧ a, a, b = a, a, x, computed in Ka,b . We compute x2 completely within Ka,b , utilizing Lemma 5.1: x2 = x1 ∨ b, a, a = a, a, x ∨ b, a, a = b, a, x = b, x, x. The element x3 = x2 ∧ q2 is computed in U , which we obtained by gluing Ka,b and B together with respect to the isomorphism ϕ1 : 1B , x, xB → 0, x, 0a,b ,

x ∈ [a, b],

16.4. 2/3-boolean triples

197

between the ﬁlter fil(q1 ) of B and the ideal Ia,b of Ka,b . So x3 is computed in two steps. First, in Ka,b : x2 ∧ va,b = b, x, x ∧ 0, b, 0 = 0, x, 0. The image of 0, x, 0 under ϕ−1 1 is b, x, x = b, x, x, so in B: x3 = b, x, x ∧ q2 = b, x, x ∧ a, b, a = a, x, a. Now comes the crucial step. To compute x4 = x3 ∨ d, c, c, we ﬁrst compute in B: x3 ∨ q3 = a, x, a ∨ a, a, b = a, x, b = x, x, b. Take the image of x, x, b under ψ and join it with d, c, c in Kc,d : x4 = x, x, bψ ∨ d, c, c = 0, xϕ, 0 ∨ d, c, c = d, xϕ, c = d, xϕ, xϕ. So x5 = x4 ∧ c, c, d = d, xϕ, xϕ ∧ c, c, d = c, c, xϕ, computed in Kc,d . To compute x6 = x5 ∨ p2 , we note that x5 ∨ vc,d = c, c, xϕ ∨ 0, d, 0 = c, d, xϕ = xϕ, d, xϕ. Then we take the image of xϕ, d, xϕc,d under ϕ2 and join it with p2 in A: x6 = xϕ, d, xϕϕ2 ∨ p2 = 0, 0, xϕ ∨ 0, 1, 0 = 0, 1, xϕ = xϕ, 1, xϕ. Finally, in A, x7 = x6 ∧ p1 = xϕ, 1, xϕ ∧ 1, 0, 0 = xϕ, 0, 0, and xϕ is identiﬁed with xϕ, 0, 0, so x7 = xϕ, as claimed. The proof for ϕ−1 is similar, using the algebraic function q(y) = ((((((y ∨p2 )∧c, c, dc,d )∨d, c, cc,d )∧q2 )∨b, a, aa,b )∧a, a, ba,b )∨a. This completes the proof of Theorem 16.1. As you can see, the four sublattices of L isomorphic to M3 play a crucial role in the proof; the elements forming these M3 -s are gray-ﬁlled in Figure 16.2.

198

16. Magic Wands

16.4.

2/3-boolean triples

To prove the surjective result, we need all the gadgets from the bijective case, and a new one. The new gadget will be described in Lemma 16.14. This new gadget is based on the 2/3-boolean triple construction, which we proceed to describe. Let N6 = {o, p, q1 , q2 , r, i} denote the six-element lattice depicted in Figure 16.3, with o the zero, i the unit element, p, q1 , q2 the atoms, satisfying the relations q1 ∨ q2 = r, p ∧ q1 = p ∧ q2 = p ∧ r = o, and p ∨ q1 = p ∨ q2 = p ∨ r = i. Following Gr¨ atzer, Greenberg, and Schmidt [32], for a bounded lattice P , we introduce 2/3-boolean triples: the element x, y, z ∈ P 3 is called a 2/3-boolean triple iff y = (y ∨ x) ∧ (y ∨ z), z = (z ∨ x) ∧ (z ∨ y). We denote by N6 [P ] the set of all 2/3-boolean triples partially ordered componentwise. We prove that N6 [P ] is a lattice and describe the congruences of this lattice. Lemma 16.3. The subset N6 [P ] of P 3 is a closure system; let x, y, z denote the closure of x, y, z ∈ P 3 and call it the 2/3-boolean closure of x, y, z. Then x, y, z = x, (y ∨ x) ∧ (y ∨ z), (z ∨ x) ∧ (z ∨ y). Proof. To simplify the notation, let y = (y∨x)∧(y∨z) and z = (z ∨x)∧(z ∨y). We have to verify that x, y, z = x, y, z is the closure of x, y, z. The triple x, y, z is 2/3-boolean closed. Indeed, y ≤ y, so x ∨ y = x ∨ y. Also, z ≤ z, so y ∨ z = y ∨ z. Therefore, (y ∨ x) ∧ (y ∨ z) = y,

i

p

r q1

q2

o Figure 16.3: The lattice N6 .

16.4. 2/3-boolean triples

199

verifying the ﬁrst half of the deﬁnition of 2/3-boolean triples; the second half is proved similarly. So x, y, z ≤ x, y, z ∈ N6 [P ]. To prove that N6 [P ] is a closure system in P 3 and that x, y, z is the closure of x, y, z, it suﬃces to verify that if x1 , y1 , z1 ∈ N6 [P ] and x, y, z ≤ x1 , y1 , z1 , then x, y, z ≤ x1 , y1 , z1 , which is obvious. Corollary 16.4. N6 [P ] is a lattice. Meet is componentwise and join is the closure of the componentwise join. Moreover, N6 [P ] has a spanning N6 (see Figure 16.4): {o = 0, 0, 0, p = 1, 0, 0, q1 = 0, 1, 0, q2 = 0, 0, 1, r = 0, 1, 1, i = 1, 1, 1}.

i

xr

xp p

r

xp

q1 xq1

q2 xq2

o Figure 16.4: Illustrating N6 [P ]. Lemma 16.5. (i) The interval [o, p] of N6 [P ] is isomorphic to P under the isomorphism xp = x, 0, 0 → x,

x ∈ P.

(ii) The interval [o, q1 ] of N6 [P ] is isomorphic to P under the isomorphism xq1 = 0, x, 0 → x,

x ∈ P.

(iii) The interval [o, q2 ] of N6 [P ] is isomorphic to P under the isomorphism xq2 = 0, 0, x → x,

x ∈ P.

(iv) The interval [p, i] of N6 [P ] is isomorphic to P under the isomorphism xp = 1, x, x → x,

x ∈ P.

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16. Magic Wands

(v) The interval [o, r] of N6 [P ] is isomorphic to P 2 under the isomorphism 0, x, y → x, y,

x, y ∈ P.

(vi) The interval [r, i] of N6 [P ] is isomorphic to P under the isomorphism xr = x, 1, 1 → x, Proof. By trivial computation. 1, x, y is closed iff x = y.

x ∈ P.

For instance, to prove (iv), observe that

For the ﬁve isomorphic copies of P in N6 [P ], we use the notation: Pp = [o, p],

Pq1 = [o, q1 ],

Pq2 = [o, q2 ],

with zero o and unit elements, p, q1 , q2 , respectively, and P p = [p, i],

P r = [r, i],

with unit i and zero elements, p, r, respectively. We describe the congruence structure of N6 [P ] based on the following decomposition of elements: Lemma 16.6. Every α ∈ N6 [P ] has a decomposition α = (α ∧ p) ∨ (α ∧ q1 ) ∨ (α ∧ q2 ), where α ∧ p ∈ Pp , α ∧ q1 ∈ Pq1 , and α ∧ q2 ∈ Pq2 . Proof. Indeed, the componentwise join of the right side equals α. For a congruence Ψ of N6 [P ], let Ψp denote the restriction of Ψ to Pp , same ˆ p denote Ψp regarded as a congruence of P ; same for for Ψq1 and Ψq2 . Let Ψ ˆ ˆ Ψq1 and Ψq2 . Similarly, let Ψp and Ψr denote the restriction of Ψ to P p and P r , ˆ p and Ψ ˆ r denote the corresponding congruences of P . respectively, and let Ψ Then we obtain: Lemma 16.7. Let α, α ∈ N6 [P ] and Ψ ∈ Con N6 [P ]. Then α ≡ α

(Ψ)

iff α ∧ p ≡ α ∧ p

(Ψp ),

α ∧ q1 ≡ α ∧ q1

Proof. This is clear from Lemma 16.6.

(Ψq1 ),

α ∧ q2 ≡ α ∧ q2

(Ψq2 ).

16.4. 2/3-boolean triples

201

Lemma 16.8. Let α = x, y, z, α = x , y , z ∈ N6 [P ] and Ψ ∈ Con N6 [P ]. Then α ≡ α (Ψ) iff x ≡ x

y ≡ y

ˆ p ), (Ψ

z ≡ z

ˆ q ), (Ψ 1

ˆ q ). (Ψ 2

Proof. This is clear from Lemma 16.7. Now we have the tools to describe the congruences. The next four lemmas provide the description. ˆq = Ψ ˆ q in P . Lemma 16.9. Let Ψ ∈ Con N6 [P ]. Then Ψ 1 2 ˆ q ), then 0, u, 0 ≡ 0, u , 0 (Ψq ), so 0, u, 0 ≡ Proof. Indeed, if u ≡ u (Ψ 1 1 0, u , 0 (Ψ). Therefore, 0, 0, u = (0, u, 0 ∨ 1, 0, 0) ∧ 0, 0, 1 ≡ (0, u , 0 ∨ 1, 0, 0) ∧ 0, 0, 1 = 0, 0, u (Ψ). ˆ q ), proving We conclude that 0, 0, u ≡ 0, 0, u (Ψq2 ), that is, u ≡ u (Ψ 2 ˆq ≤ Ψ ˆ q . By symmetry, Ψ ˆq ≥ Ψ ˆ q , so Ψ ˆq = Ψ ˆq . that Ψ 1 2 1 2 1 2 ˆp ≤ Ψ ˆq Lemma 16.10. For Ψ ∈ Con N6 [P ], the congruence inequality Ψ 2 holds. ˆ p ), then u, 0, 0 ≡ u , 0, 0 (Ψp ). Therefore, Proof. Indeed, if u ≡ u (Ψ 0, 0, u = u, 1, u ∧ 0, 0, 1 = (u, 0, 0 ∨ 0, 1, 0) ∧ 0, 0, 1 ≡ (u , 0, 0 ∨ 0, 1, 0) ∧ 0, 0, 1 = u , 1, u ∧ 0, 0, 1 = 0, 0, u (Ψ), ˆ q ), proving that Ψ ˆp ≤ Ψ ˆq . that is, u ≡ u (Ψ 2 2 Lemma 16.11. Let Θ ≤ Φ ∈ Con P . Then there is a unique Ψ ∈ Con N6 [P ], ˆ p = Θ and Ψ ˆq = Ψ ˆ q = Φ. such that Ψ 1 2 Proof. The uniqueness follows from the previous lemmas. To prove the existence, for Θ ≤ Φ ∈ Con P , deﬁne a congruence Ψ on N6 [P ] by x, y, z ≡ x , y , z (Ψ) iff x ≡ x (Θ),

y ≡ y (Φ),

z ≡ z (Φ).

202

16. Magic Wands

It is obvious that Ψ is an equivalence relation and it satisﬁes the Substitution Property for meet. To verify the Substitution Property for join, let x, y, z ≡ x , y , z (Ψ) and let u, v, w ∈ N6 [P ]. Then x, y, z ∨ u, v, w = x ∨ u, y ∨ v, z ∨ w = x ∨ u, (x ∨ y ∨ u ∨ v) ∧ (y ∨ z ∨ v ∨ w), (x ∨ z ∨ u ∨ w) ∧ (y ∨ z ∨ v ∨ w). Similarly, x , y , z ∨ u, v, w = x ∨ u, (x ∨ y ∨ u ∨ v) ∧ (y ∨ z ∨ v ∨ w), (x ∨ z ∨ u ∨ w) ∧ (y ∨ z ∨ v ∨ w). Since x ≡ x (Θ), we also have x ∨ u ≡ x ∨ u

(Θ).

From x ≡ x (Θ) and Θ ≤ Φ, it follows that x ≡ x (Φ). Also, y ≡ y (Φ), so x ∨ y ≡ x ∨ y (Φ). Therefore, x ∨ y ∨ u ∨ v ≡ x ∨ y ∨ u ∨ v (Φ). Similarly (or even simpler), y ∨ z ∨ v ∨ w ≡ y ∨ z ∨ v ∨ w (Φ). Meeting the last two congruences, we obtain that (x ∨ y ∨ u ∨ v) ∧ (y ∨ z ∨ v ∨ w) ≡ (x ∨ y ∨ u ∨ v) ∧ (y ∨ z ∨ v ∨ w)

(Φ).

Similarly, (x ∨ z ∨ u ∨ w) ∧ (y ∨ z ∨ v ∨ w) ≡ (x ∨ z ∨ u ∨ w) ∧ (y ∨ z ∨ v ∨ w)

(Φ).

The last three displayed equations verify that x, y, z ∨ u, v, w ≡ x , y , z ∨ u, v, w (Φ). Now note that for x, y ∈ P and congruence Ψ of P , xp ≡ yp

(Ψ)

iff

xr ≡ y r

xp ≡ y p

(Ψ)

iff

xq1 ≡ yq1

(Ψ)

and (Ψ).

ˆq = Ψ ˆ q and Ψ ˆr = Ψ ˆ p , so Lemmas 16.9–16.11 can be ˆp = Ψ It follows that Ψ 1 2 restated as follows: Corollary 16.12. There is a one-to-one correspondence between the congruences of N6 [P ] and pairs of congruences Θ ≤ Φ of Con P , deﬁned by ˆ r, Ψ ˆ p . Ψ → Ψ

16.4. 2/3-boolean triples

203

ˆr ≤ Ψ ˆ p can be established in a stronger form by exhibiting The inequality Ψ an algebraic function r(x) on N6 [P ] such that r(ur ) = up , for u ∈ P . Lemma 16.13. There is an algebraic function r(x) on N6 [P ] such that r(ur ) = up , for u ∈ P . Proof. Deﬁne r(x) = (((x ∧ p) ∨ q1 ) ∧ q2 ) ∨ p. Indeed, if u ∈ P , then ur = u, 1, 1 and so r(ur ) = r(u, 1, 1) = (((u, 1, 1 ∧ 1, 0, 0) ∨ 0, 1, 0) ∧ 0, 0, 1) ∨ 1, 0, 0 = ((u, 0, 0 ∨ 0, 1, 0) ∧ 0, 0, 1) ∨ 1, 0, 0 = (u, 1, 0 ∧ 0, 0, 1) ∨ 1, 0, 0 = (u, 1, u ∧ 0, 0, 1) ∨ 1, 0, 0 = 0, 0, u ∨ 1, 0, 0 = 1, 0, u = 1, u, u = up , as required. Actually, to prove Theorem 16.2, we need not the lattice N6 [P ], but a quotient thereof, which we now proceed to construct. Let P and Q be nontrivial bounded lattices and let ϕ : P → Q be a homomorphism of P onto Q. By Corollary 16.12, there is a unique congruence Ψ of N6 [P ] corresponding to the congruence pair ω ≤ ker(ϕ) of P . Deﬁne B = N6 [P ]/Ψ. This is our new gadget. It is useful to note that B can be represented as { x, y, z ∈ P × Q × Q | y = (y ∨ xϕ) ∧ (y ∨ z) and z = (z ∨ xϕ) ∧ (z ∨ y) }. By the Second Isomorphism Theorem (Theorem 1.5), there is a one-to-one correspondence between the congruences of B and congruence pairs Θ ≤ Φ of P satisfying ker(ϕ) ≤ Φ. For x ∈ N6 [P ], let x denote the congruence class x/Ψ. Using this notation, utilizing the results of this section, we state some important properties of the lattice B: Lemma 16.14. Let P and Q be bounded lattices and let ϕ : P → Q be a homomorphism of P onto Q. Then there is a lattice B, with the following properties:

204

16. Magic Wands

(i) B has a spanning sublattice o, p, q1 , q2 , r, i isomorphic to N6 . (ii) The interval [r, i] is isomorphic to P under the map x → xr , x ∈ P . (iii) The interval [p, i] is isomorphic to Q under the map y → y p , y ∈ Q, where x ∈ P with xϕ = y. (iv) The congruences Σ of B are in one-to-one correspondence with pairs of congruences Θ, Φ, where Θ is a congruence of P and Φ is a congruence of Q satisfying Θ ≤ Φϕ−1 , where up to isomorphism, Σ restricted to [r, i] is Θ and Σ restricted to [p, i] is Φ. (v) There is an algebraic function r(x) such that r(ur ) = xp , for u ∈ P , where x ∈ P with xϕ = u.

16.5.

Proof-by-Picture for surjective maps

We proceed with the Proof-by-Picture of Theorem 16.2 as in Section 16.2. We can assume, without loss of generality, that [a, b] and [c, d] are nontrivial intervals, that is, a < b and c < d.

A = M3 [K] p2

p1

p3

Fc,d

Fa,b Ka,b

Ic,d

Ia,b

Kc,d

1

r

p 0

B

Figure 16.5: The four gadgets to construct L. We take the four gadgets: A = M3 [K], Ka,b , B, and Kc,d , except that now B is not M3 [a, b] but the lattice B = N6 [a, b]/Ψ as described in Lemma 16.14, constructed from P = [a, b], Q = [c, d], and the homomorphism ϕ from [a, b] onto [c, d] given in Theorem 16.2. We do three gluings.

16.5. Proof-by-Picture for surjective maps

205

First gluing. In B, we use the ﬁlter fil(r) = { x, 1, 1 | a ≤ x ≤ b } (which is isomorphic to [a, b] since Ψ on [r, 1] is ω), while in Ka,b we utilize the ideal Ia,b = { 0, x, 0 | a ≤ x ≤ b } (which is obviously isomorphic to [a, b]), and we consider the isomorphism ϕ1 : x, 1, 1B → 0, x, 0Ka,b ,

x ∈ [a, b],

between the ﬁlter fil(r) of B and the ideal Ia,b of Ka,b to glue B and Ka,b together to obtain the lattice U . (As in Section 16.2, we use the following notation: to indicate whether a triple x, y, z belongs to A = M3 [K], Ka,b , B, or Kc,d , we subscript it: with x, y, zA , x, y, za,b , x, y, zB , or x, y, zc,d , respectively.) Second gluing. We glue Kc,d and A over the ﬁlter Fc,d = { x, c, x ∧ c | x ∈ K } of Kc,d and the ideal id(p3 ) = { 0, 0, x | x ∈ K } of A, with respect to the isomorphism ϕ2 : x, c, x ∧ cKc,d → 0, 0, xA ,

x ∈ K,

to obtain the lattice V . Final gluing. In U , we deﬁne the ﬁlter F = [p, 1] ∪ Fa,b , which is the union of [p, 1] and Fa,b , with the unit of [p, 1] identiﬁed with the zero of Fa,b . In V , we deﬁne the ideal I = Ic,d ∪ [0A , p1 ], which is the union of Ic,d and [0A , p1 ], with the unit of Ic,d identiﬁed with the zero of [0A , p1 ]. Next we set up an isomorphism ψ : F → I. Since [p, 1] = { 1, x, xB | a ≤ x ≤ b } and Ic,d = { 0, x, 0c,d | c ≤ x ≤ d },

206

16. Magic Wands

we deﬁne ψ on [p, 1] by ψ : 1, x, xB → 0, xϕ, 0c,d , where ϕ : [a, b] → [c, d] is the homomorphism given in Theorem 16.2. We deﬁne ψ on Fa,b by ψ : x, b, x ∧ ba,b → x, 0, 0A ,

x ∈ K.

It is clear that ψ : F → I is well deﬁned and it is an isomorphism. Finally, we construct the lattice L of Theorem 16.2 by gluing U over I with V over F with respect to the isomorphism ψ : F → I. The map x → x, 0, 0A is an isomorphism between K and the principal ideal id(p1 ) of A; this gives us a convex embedding of K into A. We identify K with its image, and regard K as a convex sublattice of A and therefore of L. So L is a convex extension of K. We have completed the construction of the bounded lattice L of Theorem 16.2.

16.6.

Veriﬁcation for surjective maps

The proof in Section 16.3 heavily depended on the fact that we glued over ideals and ﬁlters of which the building components were congruence-preserving extensions. This is no longer the case; however, a modiﬁcation of Lemma 2.9 comes to the rescue. A congruence Ω of L can be described by four congruences, ΩA , the restriction of Ω to A, Ωa,b , the restriction of Ω to Ka,b , Ωc,d , the restriction of Ω to Kc,d , ΩB , the restriction of Ω to B. These congruences satisfy a number of conditions: (i) 0, x, 0a,b ≡ 0, y, 0a,b (Ωa,b ) iff x, 1, 1B ≡ y, 1, 1B (ΩB ), for x, y ∈ [a, b]. (ii) 0, 0, xA ≡ 0, 0, yA (ΩA ) iff x, c, x ∧ cc,d ≡ y, c, y ∧ cc,d (Ωc,d ), for x, y ∈ K. (iii) x, 0, 0A ≡ y, 0, 0A (ΩA ) iff x, a, x ∧ aa,b ≡ y, a, y ∧ aa,b (Ωa,b ), for x, y ∈ K. (iv) 0, xϕ, 0c,d ≡ 0, yϕ, 0c,d (Ωc,d ) iff 1, x, xB ≡ 1, y, yB (ΩB ), for x, y ∈ [a, b].

16.7. Discussion

207

Conversely, if we are given congruences ΩA on A, Ωa,b on Ka,b , Ωc,d on Kc,d , ΩB on B, then by (i), we can deﬁne a congruence ΩU on U . By (ii), we can deﬁne a congruence ΩV on V . By (iii) and (iv), we can deﬁne a congruence ΩL on L. Now it is clear that if we start with a congruence Σ of K, then we can deﬁne the congruences ΣA on A, Σa,b on Ka,b , Σc,d on Kc,d componentwise, and ΣB on B as in Section 16.3. Conditions (i)–(iii) trivially hold (since ΣA , Σa,b , and Σc,d are deﬁned componentwise). Finally, (iv) holds if Σ is a ϕ-congruence. So every ϕ-congruence of K has an extension to L. Let Σ be a congruence of K that extends to L. Since A is a congruencepreserving convex extension of K = [0A , p1 ], further, Ka,b is a congruencepreserving extension of Fa,b , and Kc,d is a congruence-preserving extension of Fc,d , the congruence Σ uniquely extends to A as ΣA , to Ka,b as Σa,b , and to Kc,d as Σc,d . Therefore, Σ uniquely extends to the intervals [r, 1] and [p, 1] of B, and so by Lemma 16.14 to B. We conclude that if a congruence Σ of K extends to L, then it extends uniquely. To complete the proof, we prove that ϕ is algebraic. Deﬁne p(x) = ((((((((x ∧ a, a, ba,b ) ∨ b, a, aa,b ) ∧ p) ∨ q1 ) ∧ q2 ) ∨ d, c, cc,d ) ∧ c, c, dc,d ) ∨ p2 ) ∧ p1 . By Lemma 16.13, p(x) behaves properly in B, while outside of B, p(x) is the same algebraic function as in Section 16.3. This completes the proof of Theorem 16.2.

16.7.

Discussion

Note that Theorem 16.2 implies Theorem 16.1. However, the two theorems have diﬀerent generalizations. In [75], G. Gr¨ atzer and E. T. Schmidt proved the generalizations stated in the three subsections following. First generalization of Theorem 16.1 Let K be a bounded lattice, let [ai , bi ], i < α, be intervals of K (α is an initial ordinal ≥ 2), and let ϕi,j : [ai , bi ] → [aj , bj ],

for i, j < α,

be an isomorphism between the intervals [ai , bi ] and [aj , bj ]. For notational convenience, we write [a, b] for [a0 , b0 ]. Let Φ = { ϕi,j | i, j < α } be subject to the following conditions, for i, j < α:

208

16. Magic Wands

(1) ϕi,i is the identity map on [ai , bi ]. (2) ϕ−1 i,j = ϕj,i . (3) ϕi,j ◦ ϕj,k = ϕi,k . Let KΦ denote the partial algebra obtained from K by adding the partial operations ϕi,j , for i, j < α. Let us call a congruence Θ of K a Φcongruence iff Θ satisﬁes the Substitution Property with respect to the partial unary operations ϕ ∈ Φ. Thus a congruence relation of the partial algebra KΦ is the same as a Φ-congruence of the lattice K. We call the lattice L a Φ-congruence-preserving extension of K if a congruence of K extends to L iff it is a Φ-congruence of KΦ and every Φ-congruence of K has exactly one extension to L. Theorem 16.15. Let K be a bounded lattice, and let α ≥ 2 be an ordinal. Let [ai , bi ], for i < α, be intervals of K, and let ϕi,j : [ai , bi ] → [aj , bj ] be an isomorphism between the intervals [ai , bi ] and [aj , bj ], for i, j < α, subject to the conditions (1)–(3), where Φ = { ϕi,j | i, j < α }. Then the partial algebra KΦ has a Φ-congruence-preserving convex extension into a bounded lattice L such that all ϕi,j , i, j < α, are algebraic in L. In particular, the congruence lattice of the partial algebra KΦ is isomorphic to the congruence lattice of the lattice L. It would be nice to be able to claim that this theorem can be proved by applying Theorem 16.1 to the isomorphisms ϕi one at a time, and then forming a direct limit. Unfortunately, the direct limit at ω produces a lattice with no zero or unit, so we cannot continue with the construction. Second generalization of Theorem 16.1 Let K be a bounded lattice, let α be an ordinal, and for i < α, let ϕi be an isomorphism between the interval [ai , bi ] and the interval [ci , di ]: ϕi : [ai , bi ] → [ci , di ]. Let Φ = { ϕi | i < α }, and let KΦ denote the partial algebra obtained from K by adding the partial operations ϕi , for i < α. Let us call a congruence Θ of the lattice K a Φ-congruence iff Θ satisﬁes the Substitution Property with respect to the partial unary operations ϕi , for i < α, that is, x ≡ y (Θ) implies that xϕi ≡ yϕi (Θ), for all x, y ∈ [ai , bi ] and i < α. Thus a congruence relation of the partial algebra KΦ is the same as a Φ-congruence of the lattice K. We call the lattice L a Φ-congruence-preserving extension of the lattice K if a congruence of K extends to L iff it is a Φ-congruence of K and every Φ-congruence of K has exactly one extension to L.

16.7. Discussion

209

Theorem 16.16. Let K be a bounded lattice, let Φ be given as above. Then the partial algebra KΦ has a Φ-congruence-preserving convex extension into a lattice L such that all ϕi , for i ∈ I, are algebraic in L. In particular, the congruence lattice of the partial algebra KΦ is isomorphic to the congruence lattice of the lattice L. A generalization of Theorem 16.2 Let K be a bounded lattice, and for i ∈ I, let ϕi be a homomorphism of the interval [ai , bi ] onto the interval [ci , di ]. Let Φ = { ϕi | i ∈ I }, and let KΦ denote the partial algebra obtained from K by adding the partial operations ϕi , for i ∈ I. Let us call a congruence Θ of the lattice K a Φ-congruence iff Θ satisﬁes the Substitution Property with respect to the partial unary operations ϕi , for i ∈ I, that is, x ≡ y (Θ) implies that xϕi ≡ yϕi (Θ), for all x, y ∈ [ai , bi ] and i ∈ I. Thus a congruence relation of the partial algebra KΦ is the same as a Φ-congruence of the lattice K. We call the lattice L a Φ-congruence-preserving extension of the lattice K if a congruence of K extends to L iff it is a Φ-congruence of K and every Φ-congruence of K has exactly one extension to L. Theorem 16.17. Let K be a bounded lattice, let Φ be given as above. Then the partial algebra KΦ has a Φ-congruence-preserving convex extension into a lattice L such that all ϕi , i ∈ I, are algebraic in L. In particular, the congruence lattice of the partial algebra KΦ is isomorphic to the congruence lattice of the lattice L. Theorem 16.17 easily implies Theorems 16.15 and 16.16, with one important diﬀerence: In Theorem 16.15, we obtain a bounded lattice L. Problem 16.1. Can we ensure that the lattice L be bounded in Theorem 16.17?

Magic wands with special properties In Theorems 16.1 and 16.2, can we construct a lattice L with special properties, such as being semimodular? Obviously not if we insist on convex embeddings, since a convex sublattice of a semimodular lattice is semimodular again. Let Theorem 16.1∗ denote Theorem 16.1 with “convex” deleted, and the same for Theorem 16.2∗ . Let us call be a class C of lattices a CPE-class if every ﬁnite lattice A has a ﬁnite congruence-preserving extension B ∈ C.

210

16. Magic Wands

Theorem 16.18. Let K be a ﬁnite lattice, and let C be a CPE-class of lattices. Then in Theorems 16.1∗ and 16.2∗ , for a ﬁnite K, we can assume that L ∈ C. Problem 16.2. What can we say about CPE-classes? By Theorem 12.1, sectionally complemented lattices; by Theorem 13.1, semimodular lattices; by Theorem 14.1, isoform lattices; by Theorem 15.1, lattices with a given automorphism group form such classes. So there is a sectionally complemented magic wand, a semimodular magic wand, and so on. Problem 16.3. Is it possible to get a magic wand combining any two of these properties? Fully invariant congruences As usual, let us call a congruence Θ fully invariant iff a ≡ b (Θ) implies that aα ≡ bα (Θ), for any automorphism α. For a lattice K, let Coninv K denote the lattice of fully invariant congruences of K, and let Aut K denote the set (group) of automorphisms of K. For a bounded lattice K, we can apply Theorem 16.17 to I = Aut K; for α ∈ Aut K, let [aα , bα ] = [cα , dα ] = [0, 1], and let ϕα = α. Theorem 16.19. Let K be a bounded lattice. Then K has a convex extension into a lattice L such that a congruence of K extends to L iff it is fully invariant and a fully invariant congruence of K extends uniquely to L. In particular, Coninv K is isomorphic to Con L. The 1/3-boolean triple construction The reader may ask, what is a 1/3-boolean triple construction? For a lattice P , let us call the element x, y, z ∈ P 3 a 1/3-boolean triple iff z = (z ∨ x) ∧ (z ∨ y). Then instead of N6 of Figure 16.3, we now get the lattice N7 of Figure 16.6 (the dual of the seven-element semimodular but not modular lattice), and the 1/3-boolean triples form a lattice N7 [P ]. Problem 16.4. Is there some use of the 1/3-boolean triple construction? The Congruence Lattice Problem As discussed in the preface, the fundamental unsolved problem in this ﬁeld is the Congruence Lattice Problem: Can every distributive algebraic lattice be represented as the congruence lattice of a lattice?

16.7. Discussion

211

Figure 16.6: The lattice N7 . It is natural to ask whether the method of Schmidt [105] can be combined with the results of this paper to solve the Congruence Lattice Problem. It is pointed out in T˚ uma and Wehrung [119] that this is not the case. However, we can ask the following question. Let F0 (m) denote the free lattice with zero on m generators. Problem 16.5. Can every distributive algebraic lattice be represented as the congruence lattice of a partial algebra of the form F0 (m)Φ ? A positive answer to this question would partially answer Problem 2 of [119]. Theorems 16.1 and 16.2 and their generalizations are proved only for bounded lattices. It is easy to see, however, that with minor technical changes we can prove them for arbitrary lattices K with zero. Problem 16.6. Can we strengthen Theorems 16.1 and 16.2 to obtain a (convex) {0, 1}-extension? In Theorem 16.17, the lattice constructed may not have a zero. Problem 16.7. Can we strengthen Theorem 16.17 to obtain a bounded (convex) extension? Even stronger: a bounded (convex) {0, 1}-preserving extension?

Part V

Two Lattices

17 Sublattices

17.1.

The results

The simplest connection between two lattices K and L is the sublattice relation: K ≤ L. How then does Con K relate to Con L? As we discussed in Section 3.3, the relation K ≤ L induces a map ext of Con K into Con L: For a congruence relation Θ of K, the image ext Θ is the congruence relation of L generated by Θ, that is, ext Θ = conL (Θ). The map ext is a {0}-separating join-homomorphism. (If we want to emphasize the embedding idK : K → L, we write ext idK for ext.) In 1974, A. P. Huhn in [91] stated the converse: Theorem 17.1. Let D and E be ﬁnite distributive lattices, and let ψ : D → E be a {0}-separating join-homomorphism. Then there are ﬁnite lattices K ≤ L, and isomorphisms α : D → Con K and β : E → Con L satisfying ψβ = α(ext idK ), where idK is the embedding of K into L; that is, such that the diagram D ⏐ ⏐ ∼ =α

ψ

−−−−→

ext id

E ⏐ ⏐ ∼ =β

K Con K −−−−−→ Con L

is commutative.

216

17. Sublattices

In this chapter we prove a much stronger version of this theorem from Gr¨ atzer, Lakser, and Schmidt [46]: Theorem 17.2. Let K be a ﬁnite lattice, let E be a ﬁnite distributive lattice, and let ψ : Con K → E be a {0}-separating join-homomorphism. Then there is a ﬁnite extension L of K and an isomorphism β : E → Con L with ext idK = ψβ, that is, such that the diagram ψ

Con K −−−−→ ⏐ ⏐ =

E ⏐ ⏐ ∼ =β

ext id

K Con K −−−−−→ Con L

is commutative. While Theorem 17.1 claims the existence of the ﬁnite lattices K ≤ L, Theorem 17.2 claims that for every K, there is an extension L. This stronger form allows us to claim the existence of K and L in any CPE-class C (introduced in the subsection: Magic wands with special properties, of Section 16.7). Theorem 17.3. Let D and E be ﬁnite distributive lattices, and let ψ : D → E be a {0}-separating join-homomorphism. Let C be a CPE-class of lattices. Then there are ﬁnite lattices K, L ∈ C with K ≤ L, and isomorphisms α : D → Con K and β : E → Con L satisfying ψβ = α(ext idK ), where idK is the embedding of K into L. So there are such lattices K ≤ L that are sectionally complemented, by Theorem 12.1; semimodular, by Theorem 13.1; isoform, by Theorem 14.1; with a given automorphism group, by Theorem 15.1. Theorem 17.3 trivially follows from Theorem 17.2. Indeed, let D, E, and ψ be given as in Theorem 17.3. By the Dilworth Theorem, there is a ﬁnite lattice K1 satisfying Con K1 ∼ = D with the isomorphism δ1 . By the deﬁnition of a CPE-class, the lattice K1 has a congruence-preserving extension K ∈ C; let δ2 be an isomorphism Con K → Con K1 . Then ψ = δ2 δ1 ψ is a {0}-separating join-homomorphism of Con K into E. Applying Theorem 17.2 to K, E, and ψ , there is a ﬁnite extension L1 of K, an embedding idK : K → L1 , and an isomorphism β : E → Con L1 with ext idK = ψβ. We use once more that C is a CPE class: there is a ﬁnite congruence-preserving extension L ∈ C of L1 . Obviously, K and L satisfy the requirements of Theorem 17.3. So in the subsequent sections, we only have to prove Theorem 17.2.

17.2. Proof-by-Picture

17.2.

217

Proof-by-Picture

This Proof-by-Picture is based on the ideas in Gr¨ atzer, Lakser, and Wehrung [51] (see Section 17.5 for the history of the proofs). Let K, E, and ψ : Con K → E be given as in Theorem 17.2. We want to construct a ﬁnite extension L of K, an embedding idK : K → L, and an isomorphism β : E → Con L with ext idK = ψβ. First, let E = C2 = {0, 1}. Since ψ is a {0}-separating join-homomorphism, it follows that ωψ = 0, and Θψ = 1, for all Θ > Con K − {ωK }. By Lemma 12.3, K has a ﬁnite simple extension L, with the embedding idK : K → L. Observe that ext ωK = ωL and ext ιK = ιL , since L is simple. Deﬁne β : E → Con L by 0β = ωL and 1β = ιL . So ext idK = ψβ is obvious; it is illustrated in Figure 17.1. Second, let E = C2 × C2 . We get two projection maps πi : E → C2 , for i = 1, 2. So we can apply the E = C2 case—the last paragraph—to K, C2 , and ψπi : Con K → C2 , and we obtain the simple extensions Li of K. Obviously, L = L1 × L2 will do the job. In general, if E = Bn , we can take L = L1 × L2 × · · · × Ln , where the ﬁnite lattices L1 , L2 , . . . , Ln are simple extensions of K. Now let us take the smallest non-boolean case: E = C3 = {0, a, 1}. We embed E into B = B2 = {0, a, b, 1}, and take the lattice LB we have just

ιK

Con K

Con K − {ωK }

ψ 1 E β

ψ ωK

0

ext idK ιL

ext idK

β

Con L

ωL Figure 17.1: The case E = C2 .

218

17. Sublattices

constructed for B. Let Con LB = {ω, ΘB , ΦB , ι} and choose in LB a boolean ﬁlter F = {0F , uF , vF , 1B } so that con(uF , 1B ) = ΘB and con(vF , 1B ) = ΦB . Obviously, F is a congruence-determining sublattice of LB . By Theorem 7.9, we can represent E as the congruence lattice of a ﬁnite lattice LE with Con LE = {ω, ΘE , ι} so that LE has a boolean ideal I = {0, uE , vE , 1I } and con(0, uE ) = ΘLE and con(0, vE ) = ιLE . Obviously, I is a congruence-determining sublattice of LE . Now we construct the lattice L by gluing LB (with the ﬁlter F ) to the lattice LE (with the ideal I) with respect to the obvious isomorphism between F and I that maps uF to uI and vF to vI . In L, we use the notation u = uF = uI and v = vF = vI .

LE 1I ΘE vE

ι I ι ΘE

LE

uE

0

Θ v

1B ΘB ΦB

vF

F

ΦB ΘB

uF

Θ 0F LB

0F LB

Figure 17.2: Constructing L.

u

L

17.3. Multi-coloring

219

Since ΘB restricted to F agrees with ΘI restricted to I, it is clear that L has only one nontrivial congruence Θ, which agrees with ΘB on LB and with ΘE on LE . The maps work out with no diﬃculty (almost the same way as in Figure 17.1), verifying that L satisﬁes the requirements of Theorem 17.2. Since a formal proof of Theorem 17.2 is almost the same as that of this special case, we will not present a formal version of this Proof-by-Picture.

17.3.

Multi-coloring

In Section 3.2 we introduced colored lattices. Now we need a generalization from Gr¨ atzer, Lakser, and Schmidt [46]. Let M be a ﬁnite lattice and let C be a ﬁnite set; the elements of C will be called colors. Recall that Prime(M ) denotes the set of prime intervals of M . A multi-coloring of M over C is an map μ from Prime(M ) into Pow+ C (all nonempty subsets of C ordered by set-inclusion) satisfying the condition that if p, q ∈ Prime(M ) and con(p) ≤ con(q), then pμ ⊆ qμ. Coloring is the special case when all the pμ are singletons. Equivalently, a multi-coloring is an isotone map of the order ConJ M into the order Pow+ C. We will now show that a multi-colored lattice has a natural extension to a colored lattice. Lemma 17.4. Let M be a ﬁnite lattice with a multi-coloring μ over the set C. Then there exist a lattice M ∗ with a coloring μ∗ over C such that the following conditions holds: (i) M ∗ is the direct product of the lattices Mc , c ∈ C, where Mc is a homomorphic image of M colored by {c}. (ii) There is a lattice embedding a → a∗ of M into M ∗ . (iii) For every prime interval p = [a, b] of M , pμ = { qμ∗ | q ∈ Prime(M ∗ ) and q ⊆ [a∗ , b∗ ] } and the minimal extension of con(p) under this embedding into M ∗ is of the form (con(pc ) | c ∈ C), where pc is a prime interval of Mc iﬀ c ∈ pμ and pc is a trivial interval otherwise (in which case, con(pc ) = ωMc ). Proof. For c ∈ C, deﬁne the binary relation Φc on M as follows: u ≡ v (Φc ) iﬀ c ∈ / pμ, for every prime interval p ⊆ [u ∧ v, u ∨ v].

220

17. Sublattices

This relation is obviously reﬂexive and symmetric. To show transitivity, assume that u ≡ v (Φc ) and v ≡ w (Φc ), and let q be a prime interval in [u ∧ w, u ∨ w]. Then q is collapsed by con(u, v) ∨ con(v, w), hence there is a prime interval p in [u ∧ v, u ∨ v] or in [v ∧ w, v ∨ w] satisfying con(q) ≤ con(p). It follows from the deﬁnition of multi-coloring that qμ ⊆ pμ; since c ∈ / pμ, it follows that c ∈ / qμ, hence u ≡ w (Φc ). The proof of the Substitution Property is similar. For c ∈ C, we deﬁne the lattice Mc as M/Φc . A prime interval p of M ∗ = (Mc | c ∈ C) is uniquely associated with a c ∈ C and a prime interval of Mc . We deﬁne pμ∗ = c. It is easy to see that μ∗ is a coloring of M ∗ over C, establishing the ﬁrst condition. To establish the second condition, for a ∈ M , deﬁne a∗ so that its Mc component be a/Φc . The mapping a → a∗ is obviously a lattice homomorphism. We have to prove that it is one-to-one. Let a, b ∈ M and a = b; we have to prove that a∗ = b∗ . Let p be a prime interval in [a ∧ b, a ∨ b]. Since μ∗ is a multi-coloring, there is a c ∈ pμ∗ . Obviously, then a ≡ b (Φc ), from which the statement follows. Finally, the third condition is trivial from the deﬁnition of M ∗ and μ∗ .

17.4.

Formal proof

Now we are ready for the second proof of Theorem 17.2 as presented in Gr¨ atzer, Lakser, and Schmidt [46]. Let K, E, and ψ be given as in Theorem 17.2. Step 1. We deﬁne a map μ of Prime(K) to subsets of J(E): pμ = J(E) ∩ id(con(p)ψ). The map μ is obviously isotone. The join-homomorphism ψ separates 0, so pμ = ∅. Therefore, μ is a multi-coloring of K over J(E). We apply Lemma 17.4 to obtain the lattice K∗ = (Kc | c ∈ J(E)). Step 2. Any ﬁnite lattice M can be embedded in a ﬁnite simple lattice, Simp M , with the same zero and unit, see Lemma 12.3. Use such an extension for each Kc to obtain a simple lattice Simp Kc , then deﬁne: (Simp Kc | c ∈ J(E)), L0 = and extend the coloring so that Simp Kc is also colored by {c}. Since L0 is a direct product of simple lattices, it follows that ConJ L0 is an antichain; the congruence lattice of L0 is a boolean lattice with |J(E)| atoms. K is a

17.5. Discussion

221

sublattice of K ∗ and K ∗ is a sublattice of L0 , so we obtain an embedding ϕ : K → L0 . Finally, we construct a special ideal of L0 . Let pc be an arbitrary atom of the direct component Simp Kc ; then the prime interval [0, pc ] of L0 has color c. The atoms pc , for c ∈ J(E), generate an ideal B0 of L0 which is a boolean lattice satisfying the following properties: (1) any two distinct atoms have diﬀerent colors; (2) every color c ∈ J(E) occurs in B0 . Step 3. We continue by forming a ﬁnite atomistic lattice L1 with E ∼ = Con L1 under the isomorphism β1 . For L1 , we construct a chopped lattice P1 as in Section 7.2, except that we use a uniform “tripling”—as opposed to “doubling”—of non-maximals (for every join-irreducible element p of E, we take three atoms p1 , p2 , and p3 , so that in P1 they generate a sublattice isomorphic to M3 ). Let L1 be the ideal lattice of P1 . The isomorphism J(E) ∼ = ConJ L1 is given as follows: for p ∈ J(E), the congruence con(0, p) of L1 corresponds to p. Let β1 denote the corresponding isomorphism β1 : E → Con L1 . We consider on L1 the natural coloring over J(E) (a prime interval p is colored by con(p)β1−1 ∈ J(E)). Note that L0 and L1 are colored over the same set, J(E). Let B1 be the ideal of L1 generated by the atoms p2 , for p ∈ J(E). Then the ideal B1 is a boolean lattice satisfying the properties (1) and (2) stated in Step 2. Step 4. We have the lattice L0 with the ideal B0 and L1 with an ideal B1 . Note that B0 and B1 are isomorphic ﬁnite boolean lattices with the same coloring. Take the dual L2 of L1 ; in this lattice B1 corresponds to a ﬁlter B2 . Again, note that B0 and B2 are isomorphic ﬁnite boolean lattices with the same coloring. Glue together L0 and L2 by a color preserving identiﬁcation of B0 and B2 . The resulting lattice is L. The prime intervals of L are colored by J(E), and we have the isomorphism β : E → Con L. Since L0 is a sublattice of L, we may view ϕ as an embedding of K into L. Step 5. Finally, we have to verify that ext ϕ = ψβ. It is enough to prove that Θ(ext ϕ) = (Θψ)β, for join-irreducible congruences Θ in K. So let Θ = con(p), where p = [a, b] ∈ Prime(K). By Lemma 17.4, con(p) ext ϕ = con(a∗ , b∗ ) collapses in K ∗ the prime intervals of color ≤ Θψ; the same holds in L0 and in L. Computing (Θψ)β we get the same result, hence Θ(ext ϕ) = (Θψ)β, completing the proof of Theorem 17.2.

17.5.

Discussion

History A. P. Huhn’s 1983 paper [91]—and his later papers [92] and [93] (published posthumously)—attacked the Congruence Lattice Characterization Problem

222

17. Sublattices

(CLP) from below. He observed that a distributive algebraic lattice D with countably many compact elements is the direct limit (union) of an increasing countable family ( Di | i < ω ) of ﬁnite distributive {∨, 0}-subsemilattices of D. The Di -s are, of course, ﬁnite distributive lattices. For each i < ω, let us denote by ψi : Di → Di+1 the {∨, 0}-embedding. Huhn constructs a sequence ( Li | i < ω ) of ﬁnite lattices with lattice embeddings ϕi : Li → Li+1 such that ext ϕi : Con Li → Con Li+1 represents ψi . Then, denoting by L the direct limit of the sequence ( Li | i < ω ), he ﬁnds the representative D of L, that is, a lattice L satisfying Con L ∼ = D. The construction of the Li and ϕi is the most complicated part of his papers. However, using our Theorem 17.2, we can proceed in a straightforward manner. We ﬁrst represent D0 by a ﬁnite lattice L0 , and, inductively, given Li , we immediately get a ﬁnite lattice Li+1 and an embedding ϕi : Li → Li+1 with ext ϕi representing ψi . Pudl´ ak [100] showed that every ﬁnite subset of a distributive algebraic lattice D is contained in a ﬁnite distributive {∨, 0}-subsemilattice S of D. Of course, S is a ﬁnite distributive lattice. Pudl´ ak used this to ﬁnd a new approach to Schmidt’s result discussed in the preface. Huhn’s result also follows from Theorems 5.5 and 5.6 of Tischendorf’s 1992 thesis [117]. In 1996, G. Gr¨ atzer, H. Lakser, and E. T. Schmidt set out to give a short proof of Huhn’s result in [46]. The result published was not only a short proof but also a stronger form, with a number of applications in subsequent papers. The Proof-by-Picture we present in this chapter originated in the paper [118], in which J. T˚ uma proved the following result: Theorem 17.5. Let L0 , L1 , L2 be ﬁnite atomistic lattices and let η1 : L0 → L1 and η2 : L0 → L2 be {0}-embeddings such that ext η1 and ext η2 are injective. Let D be a ﬁnite distributive lattice, and, for i ∈ {1, 2}, let ψi : Con Li → D be {∨, 0}-embeddings such that (ext η1 )ψ1 = (ext η2 )ψ2 . Then there is a ﬁnite atomistic lattice L, and there are {0}-embeddings ϕi : Li → L, for i ∈ {1, 2}, satisfying η1 ϕ1 = η2 ϕ2 , and there is an isomorphism α : Con L → D such that (ext ϕi )α = ψi , for i ∈ {1, 2}. In Gr¨ atzer, Lakser, and Wehrung [51], T˚ uma’s result was extended: Theorem 17.6. Let L0 , L1 , L2 be lattices and let η1 : L0 → L1 and η2 : L0 → L2 be lattice homomorphisms. Let D be a ﬁnite distributive lattice, and, for i ∈ {1, 2}, let ψi : Con Li → D be complete -homomorphisms such that (ext η1 )ψ1 = (ext η2 )ψ2 . There is then a lattice L, there are lattice homomorphisms ϕi : Li → L, for i ∈ {1, 2}, with η1 ϕ1 = η2 ϕ2 , and there is an isomorphism α : Con L → D such that (ext ϕi )α = ψi , for i ∈ {1, 2}. If L0 , L1 , L2 have zero and both η1 and η2 preserve the zero, then L can be chosen to have a zero and ϕ1 and ϕ2 can be chosen to preserve the zero.

17.5. Discussion

223

If L1 and L2 are ﬁnite, then L can be chosen to be ﬁnite and atomistic. This theorem is indeed an extension of T˚ uma’s theorem—we need only observe that if the ψi are injective, then the ϕi must be lattice embeddings. This fact follows from the elementary fact that a lattice homomorphism ϕ : K → L is an embedding iff ext ϕ separates 0. Our Proof-by-Picture is the special case L0 = L1 = L2 is ﬁnite and ψ1 = ψ2 . Applications As outlined in the history subsection, for every distributive algebraic lattice D with countably many compact elements, we can ﬁnd a lattice L representing D, where L is a ω-union of ﬁnite lattices from a CPE-class C. Sometimes, we get even more as in the following two results from Gr¨atzer, Lakser, and Wehrung [51]: Theorem 17.7. Let D be a distributive algebraic lattice. If D has at most ℵ1 compact elements, then there exists a locally ﬁnite, relatively complemented lattice L with zero such that Con L ∼ = D. A lattice L is congruence-ﬁnite if Con L is ﬁnite; it is ω-congruence-ﬁnite if L can be written as a union, L = ( Ln | n < ω ), where ( Ln | n < ω ) is an increasing sequence of congruence-ﬁnite sublattices of L. We also apply Theorem 17.6 to prove the following: Theorem 17.8. Every ω-congruence-ﬁnite lattice K has a ω-congruenceﬁnite, relatively complemented congruence-preserving extension L. Furthermore, if K has a zero, then L can be taken to have the same zero. Isotone maps With reﬂections (or restrictions) we get the “dual” of Lemma 3.12. Lemma 17.9. Let K ≤ L be ﬁnite lattices. Then re : Con L → Con K is a {∧, 0, 1}-homomorphism. And we can obviously “dualize” Theorems 17.1 and 17.2. So what happens if we compose an extension and a reﬂection? Between the ﬁrst and last congruence lattices, we surely get {0}-isotone map. The converse of this was proved in Gr¨ atzer, Lakser, and Schmidt [47].

224

17. Sublattices

Theorem 17.10. Let D1 and D2 be ﬁnite distributive lattices, and let the map ψ : D1 → D2 be a {0}-isotone. Then there are ﬁnite lattices L1 , L2 , L, lattice embeddings ϕ1 : L1 → L and ϕ2 : L2 → L, and isomorphisms αi : Di → Con Li , for i = 1, 2, such that ψα2 = α1 (ext ϕ1 )(re ϕ2 ), that is, such that the diagram ψ

D1 ⏐ ⏐ ∼ =α1 ext ϕ1

Con L1 −−−−→

−−−−→

D2 ⏐ ⏐ ∼ =α2

Con L

−−−−→ Con L2

re ϕ2

is commutative. In Gr¨ atzer, Lakser, and Schmidt [50], there is a “concrete” version of this result: Theorem 17.11. Let L1 and L2 be arbitrary lattices with ﬁnite congruence lattices Con L1 and Con L2 , respectively, and let ψ : Con L1 → Con L2 be an isotone map. Then there is a lattice L with a ﬁnite congruence lattice, a lattice embedding ϕ1 : L1 → L, and a homomorphism ϕ2 : L → L2 such that ψ = (ext ϕ1 )(re ϕ2 ). Furthermore, ϕ2 is also an embedding iff ψ preserves 0. If L1 and L2 are ﬁnite, then L can be chosen to be ﬁnite and atomistic. Theorem 17.6 enabled us to prove Theorem 17.11 rather easily. Size and breadth We now have several constructions to verify Huhn’s Theorem 17.1; in all of them, the lattices K and L are very large and very “wide.” We can measure the “width” of a ﬁnite lattice L as follows. A lattice L is of breadth p if p is the smallest integer with the property that every ﬁnite X ⊆ L, there exists a Y ⊆ X such that |Y | ≤ p and for X = Y . Note that this concept is self-dual. If a ﬁnite lattice L is of breadth p, then there is an element a ∈ L with at least p covers. The breadth of the boolean lattice Bn is n. A “small” version of Theorem 17.1 was proved in [49] by G. Gr¨ atzer, H. Lakser, and E. T. Schmidt. Theorem 17.12. Let D be a ﬁnite distributive lattice with n join-irreducible elements, let E be a ﬁnite distributive lattice with m join-irreducible elements, let k = max(m, n), and let ψ : D → E be a {0}-separating join-homomorphism. Then there is a planar lattice K with O(n2 ) elements, a ﬁnite

17.5. Discussion

225

extension L ≥ K (with embedding idK : K → L) of breadth 3 with O(k 5 ) elements, and isomorphisms α : E → Con K, β : D → Con L with ψα = β(ext idK ), that is, such that the diagram D ⏐ ⏐ ∼ =β

ψ

−−−−→

E ⏐ ⏐ ∼ =α

ext id

K Con K −−−−−→ Con L

is commutative. The proof uses the planar construction of Chapter 8 and multi-coloring. Problem 17.1. Is O(k 5 ) optimal for the lattice L in Theorem 17.12. In other words, can one prove (analogously to Gr¨ atzer, Rival, and Zaguia [54]; see Chapter 8) that the size O(k 5 ) cannot be replaced by the size O(k α ), for any α < 5? Problem 17.2. Can one ﬁnd a lower bound for |L| as in Gr¨ atzer and Wang [80] and Zhang [123]? Problem 17.3. Is breadth 3 optimal for L? This is almost certainly so since a breadth 2 lattice cannot contain B3 as a sublattice, making it very diﬃcult to manipulate the congruences. However, in view of the result of the next chapter (Theorem 18.3), one can ask: Problem 17.4. Is there a planar version of Theorem 17.1? There is an interesting way of measuring the complexity of an order. The order dimension of a ﬁnite order P, ≤ is the smallest integer n ≥ 1 such that ≤ can be represented as the intersection of the orderings of n chains deﬁned on the set P . The order dimension of a ﬁnite lattice L is 1 iff L is a chain. The order dimension of L is 2 iﬀ L is planar. Problem 17.5. Prove that the lattice L we construct for Theorem 17.12 is of order dimension 3. The construction of the lattice L for Theorem 17.12 starts with the planar lattice of Chapter 8. Problem 17.6. Can one construct L for Theorem 17.12 starting from a different lattice K?

226

17. Sublattices

2-distributive lattices A. P. Huhn introduced n-distributivity in [89] and [90]. Let n ≥ 1 be an integer. A lattice L is n-distributive if for all x, y1 , . . . , yn+1 ∈ L, n+1

x∧(

i=1

yi ) =

n+1 i=1

n+1

(x ∧ (

yj )).

j=1 j=i

In particular, a lattice L is 1-distributive iff it is distributive and 2-distributive iff it satisﬁes the identity x ∧ (y1 ∨ y2 ∨ y3 ) = (x ∧ (y1 ∨ y2 )) ∨ (x ∧ (y1 ∨ y3 )) ∨ (x ∧ (y2 ∨ y3 )). We will call a lattice L doubly 2-distributive if it satisﬁes the 2-distributive identity and its dual. For instance, N5 and M3 are doubly 2-distributive lattices. In [68], G. Gr¨ atzer and E. T. Schmidt proved the following: Theorem 17.13. The lattices K and L in Theorem 17.1 can be constructed as ﬁnite doubly 2-distributive lattices. The proof is quite complex and uses multi-coloring. Problem 17.7. Can the construction for Theorem 17.13 be continued with the lattice L serving as the starting lattice? If this could be done, and repeated ω-times, then we could represent distributive algebraic lattice with countably many compact elements as congruence lattices of doubly 2-distributive lattices.

18 Ideals

18.1.

The results

The second simplest connection between two lattices K and L is that K is an ideal of L. How then does Con K relate to Con L? In Section 3.3 we discussed the connection between Con K and Con L, for the lattice L and its sublattice K. In Chapter 17, we proved the corresponding representation theorem for ﬁnite lattices. Lemma 3.11 states that if I is an ideal of the lattice L, then the restriction map re : Con L → Con I is a {0, 1}-homomorphism. The corresponding representation theorem for ﬁnite lattices was proved in G. Gr¨ atzer and Lakser [35]: Theorem 18.1. Let D and E be ﬁnite distributive lattices; let D be nontrivial. Let ϕ be a {0, 1}-homomorphism of D into E. Then there exists a ﬁnite lattice L and an ideal I of L such that D ∼ = Con I, and ϕ is represented = Con L, E ∼ by re, the restriction map. See Schmidt [111] for an alternative proof of this result. In the survey paper Gr¨ atzer and Schmidt [78], Problem 15 asks (in part), whether this result can be proved for sectionally complemented lattices. At ﬁrst glance, we may think that we can solve this problem by combining Theorem 18.1 with Theorem 12.1: Every ﬁnite lattice has a ﬁnite, sectionally complemented, congruence-preserving extension. We obtain a stronger form of Theorem 18.1: the lattice L can be assumed to be sectionally complemented. Unfortunately, in the congruence-preserving extension of L constructed in Theorem 12.1, I is no longer an ideal, so this method fails to solve this problem.

228

18. Ideals

In [42], using preorders, G. Gr¨ atzer and H. Lakser answer the question in the aﬃrmative. Theorem 18.2. Let D and E be nontrivial ﬁnite distributive lattices. Let ϕ be a {0, 1}-homomorphism of D into E. Then there exists a sectionally complemented ﬁnite lattice L and an ideal I of L such that D ∼ = Con I, = Con L, E ∼ and ϕ is represented by re, the restriction map. As an ideal of a sectionally complemented lattice, of course, the ideal I is also a sectionally complemented lattice. In Gr¨ atzer and Lakser [40], we took a diﬀerent approach to the congruence restriction problem—the strange story of how this came about is told in Section 18.6. Theorem 18.3. Let D and E be nontrivial ﬁnite distributive lattices. Let ϕ be a {0, 1}-homomorphism of D into E. Then there exists a ﬁnite planar lattice L and an ideal I of L such that D ∼ = Con I, and ϕ is = Con L, E ∼ represented by re, the restriction map. This is, of course, much stronger than Theorem 18.1: we obtain planar lattices (and very small ones). However, Theorem 18.1 is the foundation on which Theorem 18.2 is built. One cannot obtain Theorem 18.2 based on Theorem 18.3.

18.2.

Proof-by-Picture for the main result

For the Proof-by-Picture, take the ﬁve-element distributive lattices D and E and the {0, 1}-homomorphism ϕ at the top of Figure 18.1. We choose a {0, 1}homomorphism ϕ which is not an onto map. We associate with ϕ an “inverse map” : J(E) → J(D) (also shown in Figure 18.1), where on an x ∈ J(E) is deﬁned as the smallest element of D that is mapped to an element ≥ x by ϕ; see Section 2.5.2. Our plan is the following: We construct a chopped lattice M as in Section 7.2—using the gadget N6 —except that we triple, not double, every joinirreducible element, starting with the order J(D) ∪ J(E) (disjoint union). Every join-irreducible element x has three copies, xL (left), xM (middle), and xR (right). We further add two more copies of all these elements as in Section 7.2. So the new gadget (for a ∈ J(D)) is the chopped lattice G(a) in Figure 18.1, which already guarantees that aL ≡ 0 is equivalent to aM ≡ 0 is equivalent to aR ≡ 0. We take six gadgets: G(x), for each x ∈ J(D) ∪ J(E)—they are pairwise disjoint except they share the 0. We now have three chopped lattices: IC = G(d) ∪ G(e) ∪ G(i), LC = G(a) ∪ G(b) ∪ G(c), and LC = IC ∪ LC . We order the congruences in LC to agree with the ordering in J(D) as in Section 7.2, using the middle elements. For instance, to achieve that b ≡ 0

18.2. Proof-by-Picture for the main result

i

i E

D b

c

f

a

e

d

o

o

the map ϕ J(D) b

J(E) c

i

a

d

e

the map aL (aM )

aR (aL ) aL

aR

aM aL2 aM

aL1

aM (aR )

aR1

aM2

1

aR2

0 the new gadget G(a) aM (bM ) aM

aM2

bM

bM2

aM1

b M1

0

inserting N (aM , bM ) Figure 18.1: Proof-by-Picture for the main result.

229

230

18. Ideals

implies that a ≡ 0, we insert the element bM (aM ) so that N (bM , aM ) becomes a sublattice, as illustrated in Figure 18.1. So Con LC ∼ = D. Next, we order the congruences in IC to agree with the ordering in J(D). Thus Con IC ∼ = E. At this stage, there is no congruence connection between LC and IC , so Con LC is isomorphic to the direct product Con LC × Con IC ∼ = D × E. Finally, we want to achieve that x ≡ 0 is equivalent to x ≡ 0, for x ∈ J(D), for instance, that e ≡ 0 is equivalent to c ≡ 0. To accomplish this, we add the element eL (cL ) to LC so that N (eL , cL ) becomes a sublattice; therefore, in the new chopped lattice, eL ≡ 0 implies that cL ≡ 0. For the reverse implication, we add the element cR (eR ) so that N (cR , eR ) becomes a sublattice; now, cR ≡ 0 implies that eR ≡ 0. Since the gadget G(e) ensures that eL ≡ 0 is equivalent to eR ≡ 0, we conclude that eL ≡ 0 is equivalent to cL ≡ 0. We proceed similarly with all x ∈ J(D). Let LC denote the chopped lattice we obtain after adding all these elements. By construction, Con LC ∼ = D and LC contains an ideal IC with Con IC ∼ = E, and the congruences of LC restrict to IC as determined by . Now we invoke Theorem 4.6, and conclude that L = Id LC and I = Id IC satisfy the requirements of Theorem 18.1.

18.3.

A very formal proof: Main result

18.3.1

Categoric preliminaries

In our proof, we will make use of several classes of algebraic structures. It will be useful to use the language of (concrete) category theory. A concrete category is a class K of algebras (partial algebras, relational systems) called objects, along with a designated class of morphisms between any two objects. It is assumed that the identity map on an object is a designated morphism, and if K, L, M are objects, ϕ : K → L and ψ : L → M are designated morphisms, then so is ϕψ : K → M . For categories K and N, the map F : K → N is a functor if F maps objects to objects, morphisms to morphisms, and whenever ϕ : K → L and ψ : L → M are morphisms in K, then F (ϕ)F (ψ) = F (ϕψ). A contravariant functor F reverses this, that is, it satisﬁes F (ϕ)F (ψ) = F (ψϕ). Finally, we need the concept of natural equivalence. Let K and N be categories, and let F : K → N and G : K → N be functors. Then η : F → G is a natural transformation if for every object A of K, there is a morphism ηA : F (A) → G(A) such that for every morphism α : A → B in K, the diagram F (α)

F (A) −−−−→ F (B) ⏐ ⏐ ⏐ ⏐ηB ηA G(α)

G(A) −−−−→ G(B)

18.3. A very formal proof: Main result

231

is commutative, that is, F (α)ηB = ηA G(α). The functors F and G are naturally equivalent if there are natural transformations η : F → G and ν : G → F such that the map ηA νA is the identity map on F (A) and the map νA ηA is the identity map on G(A), for every object A of K. We now introduce the various categories and some of the associated functors we will use. Note that all the objects are ﬁnite structures. 1. Let DI (for DIstributive) denote the category of ﬁnite distributive lattices. The morphisms are {0, 1}-homomorphisms. 2. Let LA (for LAttice) denote the category whose objects are the ﬁnite lattices. A morphism is an ideal-embedding; see Section 1.3.2. There is a contravariant functor Con : LA → DI that associates with each lattice L its congruence lattice Con L, that is, Con(L) = Con L. If K and L are ﬁnite lattices and ε : K → L is an ideal-embedding, then Con(ε) denotes the restriction map (denoted by re in Section 3.3 and elsewhere), that is, Θ Con(ε) = ΘK , for a congruence Θ of K. By Lemma 3.11, Con(ε) is a morphism in DI . 3. Let OR (for ORder) denote the category whose objects are ﬁnite orders and whose morphisms are isotone maps. We have a contravariant functor Down : OR → DI; if ϕ : P → Q is an isotone map of orders, then the map Down ϕ : Down Q → Down P is deﬁned by H(Down ϕ) = Hϕ−1 , for each down-set H ⊆ Q. 4. Let CH (for CHopped) denote the category whose objects are chopped lattices. By an embedding ε in CH, we mean a map ε which is a one-to-one meet-homomorphism that preserves ∨ in the strong sense: xε ∨ yε exists iff x ∨ y exists and then (x ∨ y)ε = xε ∨ yε. The morphisms of CH are ideal-embeddings ε : S → T . We will show that the functor Con : CH → DI that associates with each object S of CH its congruence lattice Con S is a contravariant functor. As in the case of LA, if ε : S → T is an ideal-embedding, then Con(ε) : Con T → Con S denotes the restriction map. 5. The last category we consider is HE (for HEmi order); it is of a more technical nature. An object of HE is a ﬁnite nonempty set Q with a binary relation ρ satisfying the following three conditions: (a) ρ is irreﬂexive, that is, x ρ x fails, for all x ∈ Q; (b) ρ is antisymmetric, that is, x ρ y implies that y ρ x fails, for all x, y ∈ Q; (c) ρ is cycle-rich, that is, for each x ∈ Q, there is a y ∈ Q with y ρ x. If Q and R are objects of HE, a morphism ϕ : Q → R in HE is a one-to-one map that preserves ρ in the following strong sense: xϕ ρ yϕ

iff x ρ y.

232

18. Ideals

Let Q be an object of HE; a subset H of Q is a down-set if x ∈ H and x ρ y imply that y ∈ H. The lattice Down Q of down-sets in Q is distributive with ∅ as 0 and Q as 1. If ϕ : Q → R is a morphism in HE, then deﬁne H(Down ϕ) = Hϕ−1 , for H ∈ Down R. It is easy to see that Down ϕ : Down R → Down Q. So we have a contravariant functor Down : HE → DI. Note that we use the same symbol Down for the functors Down : OR → DI and Down : HE → DI. 18.3.2

From DI to OR

Let us recall and rephrase the results from Section 2.5.2. Given an object D of DI, that is, a ﬁnite distributive lattice, we consider the order J(D). If ϕ : D → E is a morphism in DI, then J(ϕ) : J(E) → J(D), with xJ(ϕ) = xϕ−1 , is an isotone map. Thus J is a contravariant functor J : DI → OR. Let idDI be the identity map on DI, regarded as a functor DI → DI. Note that Down J is also a functor DI → DI. Lemma 18.4. There is a natural equivalence ψ : idDI → Down J that associates with each object D of DI the isomorphism ψD : D → Down J(D) deﬁned by xψD =↓ J(D), for x ∈ D. Moreover, if D and E are ﬁnite distributive lattices and ϕ : D → E is a {0, 1}-homomorphism, then the diagram ψD

D −−−−→ Down J(D) ⏐ ⏐ ⏐ ⏐Down J(ϕ) ϕ ψE

E −−−−→ Down J(E) commutes, that is, ψD Down J(ϕ) = ϕψE , and ψD , ψE are lattice isomorphisms. 18.3.3

From OR to HE

Let P and Q be ﬁnite orders and let ϕ : Q → P be an isotone map. For P , Q, and ϕ, we construct two HE-objects, B(Q) and A(ϕ), and deﬁne the HE-morphism εϕ : B(Q) → A(ϕ); the notation reﬂects the fact that B(Q) depends only on Q, while A(ϕ) and εϕ depend on ϕ. Set B(Q) = Q × {L, M, R}; for a ∈ Q, denote the ordered pairs a, L, a, M , a, R by aL , aM , aR , respectively. Deﬁne ρ on B(Q) by setting: (α) aL ρ aM ρ aR ρ aL , for a ∈ Q; (β) aM ρ bM , whenever a b in Q, where denotes the cover relation in Q. Set A(ϕ) = (P × {L, M, R}) ∪ (Q × {L, M, R}), where we assume that P and Q are disjoint. Deﬁne ρ on A(ϕ) by setting:

18.3. A very formal proof: Main result

233

(1) aL ρ aM ρ aR ρ aL if a ∈ P ∪ Q; (2) aM ρ bM if a b, and a, b ∈ P or a, b ∈ Q; (3) aL ρ (aϕ)L , for all a ∈ Q; (4) (aϕ)R ρ aR , for all a ∈ Q. Deﬁne εϕ : B(A) → A(ϕ) by setting aK εϕ = aK , for a ∈ Q, K ∈ {L, M, R}. Then εϕ is a Q-morphism. We deﬁne the maps υ : B(Q) → Q and ϑ : A(ϕ) → P by setting aK υ = a, for a ∈ Q, K ∈ {L, M, R}; aK ϑ = aϕ, for a ∈ Q, K ∈ {L, M, R}; aK ϑ = a, for a ∈ P , K ∈ {L, M, R}. These determine the maps υ : Down Q → Down B(Q) with Hυ = Hυ −1 , for H ∈ Down Q, and ϑ : Down P → Down A(ϕ) with Hϑ = Hϑ−1 , for H ∈ Down P . The following statement is easy to verify: Lemma 18.5. The maps υ : Down Q → Down B(Q) and ϑ : Down P → Down A(ϕ) are lattice isomorphisms and the diagram ϑ

Down P −−−−→ Down A(ϕ) ⏐ ⏐ ⏐ ⏐Down ε Down ϕ ϕ υ

Down Q −−−−→ Down B(Q) commutes, that is, ϑ Down εϕ = (Down ϕ)υ . 18.3.4

From CH to DI

We need only one observation: Lemma 18.6. Con : CH → DI is a contravariant functor. Proof. Since the objects S of CH are meet-semilattices, congruence relations are determined by pairs x, y with x ≤ y. Since id(y) is a sublattice of S, we get, exactly as in the case of lattices, that for x ≤ y, the congruence x ≡ y (Θ ∨ Φ) holds iff there is a sequence x = z0 ≤ z1 ≤ · · · ≤ zn = y, with z1 ≡ zi+1 (Θ) or z1 ≡ zi+1 (Φ), for each 0 ≤ i < n. Then we can establish that Con S is a distributive lattice and that Con(ε) is a DI-morphism for any CH-morphism ε, exactly as for lattices; see Theorem 3.1.

234

18. Ideals

18.3.5

From HE to CH

We describe a functor S : HE → CH. Let Q be an object in HE and set S(Q) = {0} ∪ {a1 , a2 , a | a ∈ Q} ∪ { b(a) | a, b ∈ Q, with b ρ a }, where 0 is distinct from the other elements. Deﬁne an ordering ≤ on S(Q): 0 < ai , for i = 1, 2; ai < a, for i = 1, 2; b1 < b(a), if b ρ a; a < b(a), if b ρ a. The maximal elements of S(Q) are then of the form b(a) with b ρ a, and b, a ∈ Q, since for each a ∈ Q there is a b ∈ Q with b ρ a. Each id(b(a)) is isomorphic to the lattice N6 = N (b, a) depicted in Figure 7.1. If ϕ : Q → R is a HE-morphism, then deﬁne S(ϕ) : S(Q) → SR by setting 0S(ϕ) = 0; ai S(ϕ) = (aϕ)i ,

for i = 1, 2;

aS(ϕ) = aϕ; b(a)S(ϕ) = (bϕ)((aϕ)). Note that b ρ a implies that bϕ ρ aϕ, justifying the last equation. Lemma 18.7. S : HE → CH is a functor. Proof. In S(Q), d(c) ∧ b(a) = 0,

if b ρ a, d ρ c and a, b, c, d are all distinct;

c(b) ∧ b(a) = b1 , c(a) ∧ b(a) = a, c(a) ∧ c(b) = c1 ,

if c ρ b ρ a, and a, b, c are all distinct; if c ρ a, b ρ a, and a, b, c are all distinct; if c ρ a, c ρ b, and a, b, c are all distinct.

There appears to be one more case to check: a(b) ∧ b(a); however, this cannot occur by property (HE.b). Clearly, S(ϕ) is an ideal-embedding of S(Q) into S(R), whenever ϕ : Q → R is a HE-morphism. Given an object Q of HE, we deﬁne a map ψQ : Down Q → Con S(Q) by setting HψQ = con(H). We now prepare the proof of Theorem 18.10 with two lemmas. Given a, b ∈ Q with b ρ a, the principal ideal N (b, a) generated by b(a) has exactly three congruence relations:

18.3. A very formal proof: Main result

235

ιb,a , collapsing all of N (b, a), ωb,a , the identity relation, αb,a , depicted in Figure 7.1 (where it is denoted by Θ), with congruence classes {b1 , b(a)} and {0, a1 , a2 , a}. We ﬁrst show that ψQ is surjective. Lemma 18.8. Let Q be an object of HE and let Θ be a congruence relation on S(Q). Then H = { a ∈ Q | a ≡ 0 (Θ) } is a down-set in Q and Θ = con(H). Proof. Let b ∈ H and let b ρ a. Then b ≡ 0 (Θ), so b1 ≡ 0 (Θ). Thus ΘN (b,a) = ιb,a , and we conclude that a ∈ H. Consequently, H is a down-set in Q. Since S(Q) is sectionally complemented, Θ is determined by its congruence class containing 0. However, for any congruence, a1 ≡ 0 iff a2 ≡ 0 iff a ≡ 0, and b(a) ≡ 0 iff b ≡ 0. Thus Θ is determined by H = { a ∈ Q | a ≡ 0 (Θ) }, that is, Θ = con(H), concluding the proof. We now characterize con(H). Lemma 18.9. Let Q be an object of HE and let H be a down-set in Q. Let a, b ∈ Q with b ρ a. Then ⎧ ⎪ ⎨ιb,a , if b ∈ H; con(H)N (b,a) = αb,a , if b ∈ / H and a ∈ H; ⎪ ⎩ ωb,a , if a ∈ / H. Proof. The set Max = Max(S(Q)) of maximal elements of S(Q) consists of all elements of the form b(a), with b ρ a. For each b(a) ∈ Max, deﬁne the congruence relation Θb(a) on the ideal N (b, a) by setting

Θb(a)

⎧ ⎪ ⎨ιb,a , if b ∈ H; = αb,a , if b ∈ / H and a ∈ H; ⎪ ⎩ ωb,a , if a ∈ / H.

Then, Θb(a) N (b,a)∩N (d,c) = Θd(c) N (b,a)∩N (d,c) ,

236

18. Ideals

for any b(a), d(c) ∈ Max. Therefore, Θb(a) | b ρ a is a congruence-vector, introduced in Section 4.3, so by Lemma 4.5, there is a unique congruence relation Θ on S(Q) with ΘN (b,a) = Θb(a) ,

for all b(a) ∈ M .

By Lemma 18.9, Θ = con(H1 ),

with H1 = {a ∈ Q | a ≡ 0(Θ)}.

For each a ∈ Q, there is a b ∈ Q with b ρ a. Then a ∈ H1 iff a ≡ 0 (Θ) iff a ≡ 0 (Θb(a) ) iff a ∈ H; the last equivalence follows by the deﬁnition of Θb(a) . Thus H = H1 . We are now ready for the main result of this section. Theorem 18.10. The map ψ establishes the natural equivalence of the functors Down : HE → DI and Con S : HE → DI. That is, for every object Q in HE, ψQ : Down Q → Con S(Q) is an isomorphism and, given a HE-morphism ϕ : Q → R, the diagram in DI: ψR

Down R −−−−→ Con S(R) ⏐ ⏐ ⏐ ⏐Con S(ϕ) Down ϕ ψQ

Down Q −−−−→ Con S(Q) commutes, that is, ψR Con S(ϕ) = (Down ϕ)ψQ . Proof. We ﬁrst show that, for each object Q of HE, ψQ : Down Q → Con S(Q) is an isomorphism. Clearly, H1 ⊆ H2 implies that con(H1 ) ⊆ con(H2 ) and, by Lemma 18.8, ψQ is surjective. We need to show that ψQ is an embedding, that is, that con(H1 ) ⊆ con(H2 ) implies that H1 ⊆ H2 . Let con(H1 ) ⊆ con(H2 ) for H1 , H2 ∈ Down Q. Take a ∈ H1 , and let b ∈ Q with b ρ a; such a b exists by property (HE.c). Then a ≡ 0 (con(H2 )) and so con(H2 )N (b,a) = ωb,a . By Lemma 18.9, a ∈ H2 . Thus H1 ⊆ H2 , concluding the proof that ψQ is an isomorphism. Next we verify that the diagram is commutative. Let H ∈ Down R. Then, by Lemma 18.9, H(ψR Con S(f )) = con(H1 ),

18.3. A very formal proof: Main result

237

for some H1 ∈ Down Q. But a ∈ H1 iff a ≡ 0 (H(ψR Con S(ϕ)) iff aS(ϕ) ≡ 0 (HψR ), that is, aϕ ≡ 0 (con(H)) iff aϕ ∈ H. Thus H1 = Down Hϕ. But then H(ψR (Con S(ϕ))) = H((Down ϕ)ψQ ), and so ψR Con S(ϕ) = (Down ϕ)ψQ . 18.3.6

From CH to LA

The functor Id : CH → LA associates with each object S in CH the lattice Id(S). If ϕ : S → T is a CH-morphism, then Id(ϕ) : Id S → Id T is the idealembedding Id(I) = Iϕ. Consider the functors Con : CH → DI and Con Id : CH → DI. For any chopped lattice M , Theorem 4.6 establishes that Con M and Con Id M are isomorphic; we obtain an isomorphism by assigning to the congruence Θ of M , the congruence ΘσS = con(Θ), the congruence generated by Θ in Id M . (We regard M as a sublattice of Id M by identifying m with id(m), for m ∈ M .) Lemma 18.11. σ is a natural equivalence Con → Con Id. That is, for each object S of CH, the map σS : Con S → Con(Id S) is an isomorphism and, given a CH-morphism ϕ : S → T , the diagram σ

Con T −−−T−→ Con(Id T ) ⏐ ⏐ ⏐ ⏐Con(Id(ϕ)) Con(ϕ) σ

Con S −−−S−→ Con(Id S) in DI commutes, that is, σT Con(Id(ϕ)) = (Con(ϕ))σS . 18.3.7

The ﬁnal step

By combining the natural equivalences of Lemmas 18.4, 18.11, and Theorem 18.10, and the commutative diagram of Lemma 18.5, we get our main result: Theorem 18.12. Let D and E be ﬁnite distributive lattices and let ϕ : D → E be a {0, 1}-homomorphism. Then there exist ﬁnite lattices I, L, an ideal-embedding α of I into L, and isomorphisms β : D → Con L, γ : E → Con I such that the diagram β D −−−−→ Con L ⏐ ⏐ ⏐ ⏐Con(ϕ) ϕ γ

E −−−−→ Con I

238

18. Ideals

is commutative, that is, β Con(ϕ) = ϕγ. Proof. Consider the following diagram: ψD

ϑ

ϕA(J(ϕ))

ψE

u

ϕB(J(E))

D −−−−→ Down J(D) −−−−→ Down A(J(ϕ)) −−−−−→ . . . ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ϕ Down εJ(ϕ) Down J(ϕ) E −−−−→ Down J(E) −−−−→ Down B(J(E)) −−−−−→ . . . ϕA(J(ϕ))

σS(A)(J(ϕ))

ϕB(J(E)))

σS(B)(J(E))

. . . −−−−−→ Con S(A)(J(ϕ)) −−−−−−−→ Con Id S(A)(J(ϕ)) ⏐ ⏐ ⏐ ⏐ Con SεJ(ϕ) Con Id S(A)(J(ϕ)) . . . −−−−−−→ Con S(B)(J(E)) −−−−−−−→ Con Id S(B)(J(ϕ)) By Lemma 18.4, ψD and ψE are isomorphisms and the left-most square commutes. By Lemma 18.5, the maps u and ϑ are isomorphisms and the next square commutes. By Theorem 18.10, ψA(J(ϕ)) and ψB(J(E)) are isomorphisms and the corresponding square commutes. Finally, by Lemma 18.11, ωS(A)(J(ϕ)) and ωS(B)(J(E)) are isomorphisms and the right-most square commutes. Set L = Id S(A)(J(ϕ)), I = Id S(B)P (J(E)), α = Id S(εJ(ϕ) )). If we set β = ψD ϑ ψA(J(ϕ)) σS(A)(J(ϕ)) and

γ = ψE υ ψB(J(E)) σS(B)(J(E)) ,

the theorem follows.

18.4.

Proof for sectionally complemented lattices

We constructed a lattice L for Theorem 18.1 in Section 18.3.7. To verify Theorem 18.2, we have to prove that L is sectionally complemented. We will do this in Theorem 18.13 in a very general setup. Let Q be a ﬁnite preorder with the covering relation and let P be the ﬁnite order Q/≈ associated with it; see Section 1.1.4. An element q of Q is said to be maximal if there is no p ∈ Q with p q. Let Max(Q) be the set of maximal elements in Q. We deﬁne a set T consisting of one element p1 = p2 , for each maximal p ∈ P , and of two distinct elements p1 , p2 , for each nonmaximal p ∈ P . We work with the subscripts modulo 2, so p3 = p1 , for any p ∈ P . A subset I of T is said to be closed if (Cl)

p1 , qi ∈ I implies that qi+1 ∈ I, for p q in P .

The closure of a subset X of T is denoted X. The set of closed subsets of T forms a lattice L, whereby meet is set intersection and the join of two closed subsets I and J is given by I ∨ J = I ∪ J.

18.4. Proof for sectionally complemented lattices

239

Theorem 18.13. The lattice L of closed subsets of T is sectionally complemented. Let I, J be closed subsets of T with I ⊆ J. We construct a sectional complement of I in J. We ﬁrst deﬁne several subsets of P . Deﬁne S by setting (1)

S = { p ∈ P | pi ∈ J − I, for some i = 1, 2 },

and partition S into two disjoint subsets S0 and S1 by setting (2)

/ I, for all i = 1, 2 } S 1 = { p ∈ S | pi ∈

and (3)

S0 = S − S1 = { p ∈ S | pi ∈ J − I, pi+1 ∈ I, for some i = 1, 2 }.

We restrict to S0 and let ≥ be the reﬂexive-transitive closure of restricted to S0 . That is, if p, q ∈ S0 , then p ≥ q iff either p = q or there is a k > 0 and a0 , a1 , . . . , ak ∈ S0 with p = a0 a1 · · · ak = q. We get an equivalence relation ≈ on S0 whereby p ≈ q iff p ≥ q and q ≥ p, and the set of equivalence classes S0 is a ﬁnite order under the induced ordering, which we will also denote by ≥. We denote the equivalence class of ≈ containing p ∈ S0 by p/≈. We ﬁrst present two lemmas useful for our calculations. Lemma 18.14. Let p ∈ S0 and let there be an element q ∈ S0 distinct from p with q ≈ p. Then p1 ∈ J − I and p2 ∈ I. Proof. By (3), exactly one of p1 , p2 is in I and the other is in J − I. Assume, to the contrary, that p1 ∈ I. Since p and q are equivalent and distinct, there is a k > 0 and a0 , . . . , ak in S0 with p = a0 a1 · · · ak = q. Now p1 = a01 ∈ I, one of a11 , a12 ∈ I, and I is closed. Thus both a11 , a12 ∈ I, contradicting the fact that a1 ∈ S0 . Consequently, p1 ∈ J − I and so p2 ∈ I, by (3). Lemma 18.15. Let p ∈ S0 , let p1 ∈ J − I, and let q ∈ S0 satisfy q ≈ p. Then qi ∈ I ∪ {p1 }, for i = 1, 2. Proof. Since p ∈ S0 , it follows that p2 ∈ I, and so the conclusion holds if q = p. Otherwise, there are k > 0 and a0 , . . . , ak ∈ S0 with p = a0 a1 · · · ak = q. Since q ≈ p, we have ai ≈ p, for all i = 0, . . . , k. Thus, by Lemma 18.14, a11 , a21 , . . . , ak1 = q1 ∈ J − I

240

18. Ideals

and a12 , a22 , . . . , ak2 = q2 ∈ I. It follows that p1 ∈ I ∪ {p1 } and a12 ∈ I ⊆ I ∪ {p1 }. Thus a11 ∈ I ∪ {p1 }. Since a22 ∈ I ⊆ I ∪ {p1 }, we conclude that a21 ∈ I ∪ {p1 }. Proceeding in this manner, we ﬁnally conclude that q1 = ak1 ∈ I ∪ {p1 }. We now construct a complement Y of I in J. We deﬁne a subset X (the set of bad cycles) of S0 : An equivalence class β ∈ X iff there is b ∈ β and a ∈ S with a b, with a1 ∈ J − I, and with either a ∈ S1 or a ∈ S0 with a/≈ = β. Set Y = S0 −X and let Y0 ⊆ S0 be a transversal of Y , that is, |Y0 ∩α| = 1, for each α ∈ Y and Y0 ∩ β = ∅, for each β ∈ X . We now deﬁne Y ⊆ J by setting (4)

Y = { pi ∈ J − I | p ∈ Y0 } ∪ { pi ∈ J | p ∈ S1 }. We show that Y is the desired complement of I in J. By (2), Y ⊆ J − I.

Lemma 18.16. Y is closed. Proof. Let a b, let a1 ∈ Y , and let bi ∈ Y , for some i. Either b ∈ S0 or b ∈ S1 . First, assume that b ∈ S0 . Then a ∈ S0 ; indeed, otherwise, a ∈ S1 and / Y , so, by the deﬁnition of X , we would get b/ ≈ ∈ X and thus b/ ≈ ∈ contradicting that bi ∈ Y . It follows that a/ ≈ = b/ ≈, since otherwise, b/≈ ∈ X again. Since is irreﬂexive, it follows that a = b, and by deﬁnition of Y0 , we cannot have both a ∈ Y0 and b ∈ Y0 , that is, we cannot have both a1 ∈ Y and bi ∈ Y . So we can assume that b ∈ S1 . Since J is closed, b1 , b2 ∈ J, and so we conclude that b1 , b2 ∈ Y , that is, bi+1 ∈ Y , by (4), the deﬁnition of Y . In order to show that I ∨ Y = J, we ﬁrst prove the following lemma. Lemma 18.17. If pi ∈ J − I ∪ Y , then p ∈ S0 and p/≈ ∈ X . Proof. Clearly, p ∈ S0 , for otherwise p ∈ S1 and so pi ∈ Y by (4). Thus pi+1 ∈ I, by (3), the deﬁnition of S0 . Therefore, either p/≈ ∈ Y or p/≈ ∈ X . But if p/ ≈ ∈ Y , then there is an a ∈ S0 with a ≈ p and a ∈ Y0 . If a = p, then, since ai+1 = pi+1 ∈ I, pi = ai ∈ Y ⊆ I ∪ Y . Thus a = p. Then, by Lemma 18.14, a1 ∈ J − I, that is, a1 ∈ Y . But, by Lemma 18.15, then pi ∈ I ∪ {a1 } ⊆ I ∪ Y . Consequently, p/≈ ∈ X . Lemma 18.18. J = I ∨ Y .

18.5. Proof-by-Picture for planar lattices

241

Proof. Assume, to the contrary, that J = I ∨ Y = I ∪ Y . Then there is a p ∈ S with pi ∈ J − I ∪ Y . By Lemma 18.17, p ∈ S0 and p/≈ ∈ X . Since X is a ﬁnite order, there is then a p ∈ S0 such that p/≈ is maximal (under the order ≥) with respect to the property: pi ∈ / J − I ∪ Y , for some i = 1, 2. That is, if q/≈ ∈ X and q/≈ > p/≈, then q1 , q2 ∈ I ∪ Y . Take such a pi . Now, p ∈ S0 and pi ∈ / I. So pi+1 ∈ I. Since p/≈ ∈ X , there is a q ∈ S0 with q ≈ p, there is an a ∈ S with a1 ∈ J − I, with a q, and with either a ∈ S1 or a ∈ S0 and a/≈ = q/≈ = p/≈. / I ∪ Y , for otherwise, by Lemmas 18.14 and 18.15, If q = p, then q1 ∈ q1 ∈ J − I and pi ∈ I ∪ {q1 } ⊆ I ∪ Y . Thus we can replace p by q in our considerations, that is, we may assume that a p. Now if a ∈ S1 , then a1 ∈ Y ⊆ I ∪ Y and pi+1 ∈ I ⊆ I ∪ Y , implying the contradiction pi ∈ I ∪ Y . On the other hand, if a ∈ S0 , then a/≈ = p/≈ and a p imply that a/≈ > p/≈. If a/≈ ∈ X , then, by the maximality of p/≈, we obtain that a1 ∈ I ∪ Y and, since pi+1 ∈ I ⊆ I ∪ Y , we get the contradiction pi ∈ I ∪ Y . Thus a/≈ ∈ Y . So there is a b ∈ a/≈ with bi ∈ Y , for some i. However, i = 1, for if b = a, then, by Lemma 18.14, b1 ∈ J − I, and if b = a, then b1 = a1 ∈ J − I. Using Lemma 18.15 for b = a, we conclude that a1 ∈ I ∪ {b1 } ⊆ I ∪ Y . Since pi+1 ∈ I ⊆ I ∪ Y , we again get the contradiction pi ∈ I ∪ Y . Thus the assumption J = I ∨ Y leads to a contradiction, proving the lemma. Thus the lattice L is sectionally complemented, proving Theorem 18.13.

18.5.

Proof-by-Picture for planar lattices

The Proof-by-Picture will use the example distributive lattices D and E and the {0, 1}-homomorphism ϕ of Figure 18.2. The ﬁgure also shows the “inverse map” : J(E) → J(D), as it was done in Section 18.2; see also Section 2.5.2. As in Section 8.2, we base the construction on the direct product of two chains C and C . Again we use two gadgets: covering M3 -s to make a prime interval p of the ﬁrst chain congruence-equivalent to a prime interval q of the second chain; and N5 (rather than the N5,5 of Section 18.2) to force that con(p) < con(q), for the prime intervals p and q of the ﬁrst chain. Let C = C7 and C = C6 . For every cover x ≺ y in J(D) ∪ J(E), we color three adjacent prime intervals of C with y, x, and y. We start with J(E); it has one cover: d ≺ i, so we color the ﬁrst three prime intervals with i, d, i. Then we take the only cover of J(E): b ≺ c, and we color the next three prime intervals of C with b, b, c. We color C by setting up a bijection between the prime intervals of C and J(D) ∪ J(E). We start with J(E), then with J(D). We start by adding the M3 gadgets.

242

18. Ideals

i i E

D c d b

a

o

o

ϕ: D → E

a

J(D) c

J(E) i

b

d

: J(E) → J(D) Figure 18.2: Distributive lattices for the planar example. First, the M3 gadget is used to “ﬁll in” every covering square whose two sides have the same color; there are six. Second, we identify by . We “ﬁll in” every covering square whose two sides have color x and x, for some x ∈ J(E); there are two. Next comes the N5 gadget: for every cover x ≺ y in J(D) ∪ J(E), take the three adjacent prime intervals of C colored with y, x, and y; add an element so that they will form an N5 . Note that the prime interval colored by a is not related by any other prime interval by the N5 or M3 gadget because a is not comparable to any other element of J(D) and because a is not in the image of . This completes the construction of the lattice L. The principal ideal generated by the black-ﬁlled element is I; see Figure 18.3. It is really easy to compute that I represents E, L represents D, and the restriction map represents ϕ. A formal proof Theorem 18.3 is very similar to proofs we have already presented in detail, so we leave it to the reader.

18.6.

Discussion

In general, automorphisms of a lattice do not restrict to automorphisms of an ideal. In [40], G. Gr¨ atzer and H. Lakser construct lattices where this does happen.

18.6. Discussion

243

c c

b c

b i

a d

i i

d

Figure 18.3: The planar lattice L and ideal I. Theorem 18.19. Let D and E be ﬁnite distributive lattices with more than one element, and let ψ : D → E be a {0, 1}-homomorphism. Let G and H be groups, and let η : G → H be a group homomorphism. Then there exist a lattice L, an ideal I in L, lattice isomorphisms D : D → Con L,

E : E → Con I,

and group isomorphisms τG : G → Aut L,

τH : H → Aut I

such that, for each x ∈ D, the congruence relation xψE on I is the restriction to I of the congruence relation xD on L, and, for each g ∈ G, the automorphism gητH of I is the restriction of the automorphism gτG of L. If G and H are ﬁnite, then the lattice L can be chosen to be ﬁnite.

244

18. Ideals

By identifying D with Con L, E with Con I, G with Aut L, and H with Aut I, Theorem 18.19 can be paraphrased as follows: any pair ψ, a {0, 1}homomorphism of ﬁnite distributive lattices, and , a homomorphism of groups, can be simultaneously realized as the respective restrictions Con L → Con I and Aut L → Aut I, for some lattice L and some ideal I in L. Note that this result is a far reaching generalization of the Baranski˘ı-Urquhart Theorem discussed in Section 15.1. In Section 17.5 we discussed the (doubly) 2-distributive lattices of A. P. Huhn. Problem 18.1. Can Theorem 18.1 be proved for (doubly) 2-distributive lattices? Problem 18.2. What can we say about semimodular (doubly) 2-distributive lattices? Problem 18.3. Can Theorem 18.1 be proved for (semi) modular lattices?

19 Tensor Extensions

19.1.

The problem

Let A and B be nontrivial ﬁnite lattices. Then Con A × Con B is a ﬁnite distributive lattice, so by the Dilworth Theorem (Theorem 7.1), the lattice Con A × Con B can be represented as Con L, for some ﬁnite lattice L. How can we construct the lattice L from the lattices A and B? By Theorem 2.1, (1)

Con(A × B) ∼ = Con A × Con B,

so we can take L = A × B. For a similar isomorphism, we now introduce tensor products. For a nontrivial lattice K with zero, we denote by K − the ∨-subsemilattice of K deﬁned on K − {0}. Let A and B be nontrivial lattices with zero. We denote by A ⊗ B the tensor product of A and B, deﬁned as the free {∨, 0}-semilattice generated by the set A− × B − and subject to the relations (2)

a, b1 ∨ a, b2 = a, b1 ∨ b2 ,

for a ∈ A− , b1 , b2 ∈ B − ;

a1 , b ∨ a2 , b = a1 ∨ a2 , b,

for a1 , a2 ∈ A− , b ∈ B − .

If A and B are ﬁnite nontrivial lattices, then A⊗B is a ﬁnite lattice since it is a ﬁnite ∨-semilattice with zero. Since A ⊗ B just “happens to be a lattice”, it is quite surprising that the isomorphism (3)

Con A ⊗ Con B ∼ = Con(A ⊗ B),

246

19. Tensor Extensions

holds, see Gr¨ atzer, Lakser, and Quackenbush [43]. The proof of this result is very complicated, and not appropriate for presentation in this book. There is, however, a closely related result, with an approachable proof. If L is a ﬁnite lattice and D is a ﬁnite distributive lattice, then we can deﬁne L[D] as the lattice of all isotone maps from J(D) to L. By Schmidt [108], we obtain that Con L[D] ∼ = (Con L)[Con D]. (See also Section 5.3 and Gr¨ atzer and Schmidt [61].) So here is the question: Let A and B be nontrivial ﬁnite lattices. Then (Con A)[Con B] is a ﬁnite distributive lattice. Can we represent it as Con L, for a ﬁnite lattice L constructed from A and B? In Gr¨ atzer and Greenberg [29], we introduced a construction: the tensor extension A[B], for nontrivial ﬁnite lattices A and B. In this chapter we will prove that (4)

Con(A[B]) ∼ = (Con A)[Con B],

providing an answer to the above question. In Section 19.7 we discuss in detail the history of this isomorphism and how it relates to tensor products. In [29], we prove that for ﬁnite lattices, a tensor extension is isomorphic to a lattice tensor product, introduced in Gr¨ atzer and Wehrung [82], so our result is the ﬁnite case of the main result of [82]. The reader should view this chapter as an introduction to a very technical chapter of congruence lattice theory, and should consult Section 19.7 for further readings. This elementary introduction is based on Gr¨ atzer and Greenberg [29] and [31].

19.2.

Three unary functions

To prepare for the introduction of tensor extensions, we deﬁne some useful functions. Let A and B be nontrivial ﬁnite lattices. We are going to deﬁne three unary functions, m, j, and p, on B A , the set of all maps from B into A. First, for a ∈ A and α ∈ B A , we deﬁne—computed

19.2. Three unary functions

247

B

A

α a

(A − fil(a))α

A − fil(a)

ma (α) =

(A − fil(a))α

B A ja (α) =

(A − id(a))α

A − id(a)

α

a

(A − id(a))α

Figure 19.1: Illustrating ma (α) and ja (α). in B (see Figure 19.1): ma (α) = (A − fil(a))α = (5) xα, (6) (7)

ja (α) = pa (α) =

xa

(A − id(a))α =

xa

yα,

ya

jx (α) =

xa

(A − id(x))α =

xa yx

yα.

248

19. Tensor Extensions

The following statements about these functions are easy to verify: Lemma 19.1. For a, b ∈ A and α ∈ B A , ma (α) ∧ mb (α) = ma∨b (α),

(8) (9)

ja (α) ∨ jb (α) = ja∧b (α).

Moreover, m0 (α) = 1B , j1 (α) = 0B , and p0 (α) = 1B . For A = M3 = {0, a, b, c, 1}, we get ma (α) = bα ∧ cα ∧ 0α, and symmetrically for mb (α) and mc (α). These are the expressions that occur in the existential deﬁnition of boolean triples. For A = M3 , pa (α) = jb (α) ∧ jc (α) ∧ j0 (α) = (aα ∨ cα ∨ 1α) ∧ (bα ∨ cα ∨ 1α) ∧ (aα ∨ bα ∨ cα ∨ 1α) = (aα ∨ cα ∨ 1α) ∧ (bα ∨ cα ∨ 1α), and symmetrically for pb (α) and pc (α). These are the same expressions that occur in the ﬁxed-point deﬁnition of boolean triples. Now we deﬁne the function m : B A → B A by (10)

am(α) = ma (α),

for a ∈ A and α ∈ B A . We similarly deﬁne the functions j and p: (11) (12)

aj(α) = ja (α), ap(α) = pa (α).

So, by deﬁnition, p(α) = m(j(α)), or even simpler, (13)

19.3.

p = m ◦ j.

Deﬁning tensor extensions

In Section 5.1 we had three deﬁnitions of M3 [B]: the “ﬁxed point deﬁnition”, (F), the “existential deﬁnition”, (E), and the “closure deﬁnition”, Lemma 5.1.(ii). The elements of M3 [B] are 3-tuples of elements of B. Similarly, the elements of the tensor extension A[B] are deﬁned as A-tuples of elements of B that is, elements of B A . Let A and B be ﬁnite nontrivial lattices. We deﬁne (the existential deﬁnition) the tensor extension of A by B: (14)

A[B] = { m(α) | α ∈ B A }.

19.3. Deﬁning tensor extensions

249

In other words, for γ ∈ B A , γ ∈ A[B]

there exists α ∈ B A satisfying γ = m(α).

iff

To obtain a “ﬁxed-point description”, we introduce the notation p(α) = α (for α ∈ B A ) and call α the closure of α (in B A ). We now verify that this is indeed a closure operation on B A . Lemma 19.2. For α, β ∈ B A , (i) α ≤ α. (ii) If α ≤ β and β ∈ A[B], then α ≤ β. (iii) α ∈ A[B] iff α = α. Proof. (i)Let a ∈ A. Then for any x a, we have a ∈ A − id(x), and so aα ≤ α(A − id(x)). Therefore, by (7), aα ≤ α(A − id(x)) = pa (α) = aα. xa

(ii) By the deﬁnition of A[B], there is a γ ∈ B A such that β = m(γ). Choose a ∈ A and x a. Thus for each y ∈ A − id(x), we have yα ≤ yβ = my (γ) =

(A − fil(y))γ ≤ xγ,

since x ∈ A − fil(y). Joining these inequalities for y ∈ A − id(x), we obtain that (A − id(x))α ≤ xγ. Meeting these inequalities for x a, we get pa (α) =

(A − id(x))α ≤

xa

xγ = ma (γ) = aβ,

xa

or aα ≤ aβ, proving that α ≤ β. A (iii) Let α = α, that is, aα = pa (α), for all a ∈ A. Deﬁne β ∈ B by bβ = (A − id(b))α, for b ∈ A. Then ma (β) = (A − fil(a))β = (A − id(x))α = pa (α) = aα, xβ = xa

xa

for any a ∈ A; therefore, α ∈ A[B]. Conversely, assume that α ∈ A[B]. By (i), α ≤ α holds. The reverse inequality holds by (ii) with β = α, so α = α.

250

19. Tensor Extensions

Lemma 18.15 shows that A[B] is a closure system in B A . Thus we have established the following result. Theorem 19.3. Let A and B be nontrivial ﬁnite lattices. The tensor extension A[B] is a lattice with meets computed pointwise and joins computed as the closures of the pointwise joins, that is, according to the formula α ∨A[B] β = α ∨B A β, for α, β ∈ A[B]. The main result on ﬁnite tensor extensions is the following: Theorem 19.4. Let A and B be nontrivial ﬁnite lattices. Then the following isomorphism holds: Con(A[B]) ∼ = (Con A)[Con B]. If A is simple, then Con A ∼ = C2 , so (Con A)[Con B] ∼ = Con B. We have obtained the following generalization of Theorem 5.3. Corollary 19.5. Let A and B be nontrivial ﬁnite lattices. If A is simple, then A[B] is a congruence-preserving extension of B.

19.4.

Computing

19.4.1

Some special elements

Let A and B be nontrivial ﬁnite lattices. For any q ∈ B, we deﬁne the (almost) constant function, κq , in B A with value q: q, for x ∈ A − {0A }; κq (x) = 1B , for x = 0A . Clearly κq ∈ A[B], for any q ∈ B, since p(κq ) = κq . We will call { κq | q ∈ B } the diagonal of A[B]. For a ∈ A, we deﬁne the element χa ∈ B A —the characteristic function of id(a)—by (see Figure 19.2) 1B , if x ∈ id(a); χa (x) = 0B , otherwise. Lemma 19.6. χa is an element of A[B], for any a ∈ A. Proof. Deﬁne γ ∈ B A by fil(a)γ = 0B and (A − fil(a))γ = 1B . Clearly, mb (γ) = bχa , for any b ∈ A, that is, χa = m(γ), so χa ∈ A[B].

251

19.4. Computing

A

0B

A

b

0B

pab

χa a

p a

1B

1B

Figure 19.2: Illustrating χa and pab . If p ∈ B and a, b ∈ A with a < b, we deﬁne pab = χb ∧ (χa ∨B A κp ) ∈ B A (see Figure 19.2). Lemma 19.7. pab ∈ A[B]. Proof. Under these hypotheses, χa ∨B A κp = χa ∨A[B] κp ∈ A[B], therefore, pab ∈ A[B]. The proof of the following statement is trivial since the joins and the relevant meets in A[B] are computed pointwise. Lemma 19.8. Let a, b ∈ A with a < b. Then the map ϕab : p → pab is an embedding of B into the sublattice [χa , χb ] of A[B]; in fact, ϕab is a {0, 1}embedding. Proof. This is obvious. The last statement adds the facts that (0B )ab = χa and (1B )ab = χb . We will denote by Bab the image of B under the map ϕab . Note that the diagonal, { κq | q ∈ B }, can also be deﬁned as B0A 1A . Moreover, if a is covered by b and b is join-irreducible, then id(b) − id(a) = {b}, so Bab = [χa , χb ]. The next lemma will be useful in computing some joins in A[B]. Lemma 19.9. Let a, b, c ∈ Deﬁne α ∈ B A by ⎧ 1B , ⎪ ⎪ ⎪ ⎨ q, tα = ⎪ p, ⎪ ⎪ ⎩ 0B ,

A with c = a ∨ b and let p, q ∈ B with p ≤ q. for for for for

Then α = qac . (See Figure 19.3.)

t ∈ id(a); t ∈ id(b) − id(a); t ∈ id(c) − (id(a) ∪ id(b)); t ∈ A − id(c).

252

19. Tensor Extensions

A

0B

0B

c

A

c p

a

q

a

b

b

q 1B

1B α

α = qac

Figure 19.3: Illustrating α and α. Proof. Since α ≤ qac , it follows that α ≤ qac . So it is suﬃcient to prove that α(t) = q, for t ∈ id(c)−(id(a)∪id(b)). To see this, let e ∈ id(c)−(id(a)∪id(b)), that is, e ≤ c, but e a and e b. Let f ∈ A − fil(e), that is, e f . Then either a f or b f ; otherwise, we would get a ∨ b = c ≤ f , implying that e ≤ f , a contradiction. Therefore, α(A − id(f )) ≥ aα ∧ bα = 1B ∧ q = q, from which it follows that eα =

α(A − id(f )) ≥ q,

f ∈A−fil(e)

for any e ∈ id(c) − (id(a) ∪ id(b)). Since α(e) ≤ qac , by Lemma 18.15.(ii), eα ≤ qac , which implies that eα ≤ q. We conclude that eα = q, as claimed. Lemma 19.10. The map a → χa is an embedding of A into A[B]. Proof. This map is clearly one-to-one and meet-preserving. To compute χa ∨ χb , let α be the pointwise join. Applying Lemma 19.9, we obtain that α = χa∨b , proving that χa ∨ χb = χa∨b . Note that while we get many copies of B in A[B], we obtain only one copy of A in A[B]—this is not surprising since there is, in general, only one copy of M3 in M3 [B] and, further, A[C2 ] is isomorphic to A. 19.4.2

An embedding

Let A and B be ﬁnite nontrivial lattices. Then for an ideal I of A, we have a natural embedding of I[B] into A[B].

19.4. Computing

253

Lemma 19.11. Let I be an ideal of A. Then there exists a unique embedding f : I[B] → A[B], such that f ( pab ) = pab , for p ∈ B and a < b in I, where in the formula f ( pab ) = pab , the ﬁrst pab is in I[B], while the second is in A[B]. Proof. For α ∈ I[B], we deﬁne αz (∈ B A ), the “zero-padded” version of α, by xα, if x ∈ I; xαz = 0B , otherwise. It is obvious that αz ∈ A[B]. Deﬁne f : I[B] → A[B], by α → αz . Clearly, f is an embedding and f ( pab ) = pab . By Lemma 19.22, we have the uniqueness. 19.4.3

Distributive lattices

For the lattices X and Y , let hom{∨,0} (X, Y ) denote the set of {∨, 0}-homomorphism of X into Y , regarded as a {∨, 0}-semilattice. The following statement is a generalization of the fact that balanced triples are boolean for a distributive lattice B; see Section 5.3. Lemma 19.12. Let B be a distributive lattice. Then A[B] = hom{∨,0} (A, B d ).

(15)

Proof. By Lemma 19.1, for any lattice B, we always have A[B] ⊆ hom{∨,0} (A, B d ). To prove the reverse containment in (15), let us assume that B is distributive. Let α ∈ hom{∨,0} (A, B d ), and let p be the polynomial introduced in (7). By Lemma 19.2, it is suﬃcient to show that p(α) ≤ α. For a ∈ A, compute: ap(α) =

yα

xa yx

=

( xνα | ν ∈ (A − id(x)) ) xa

=

xν)α | ν ∈ (A − id(x) )α ( xa

(by the distributivity of B)

xa

xa

(since α ∈ hom{∨,0} (A, B d )).

254

19. Tensor Extensions

To show that p(α) ≤ α, that is, that ap(α) ≤ aα, for all a ∈ A, it suﬃces to show that (16) ( xν)α ≤ aα, xa

for any ν ∈ xa (A − id(x)). Let ν ∈ xa (A − id(x)), and let b = xa xν. We claim that b ≥ a. Indeed, otherwise, fromthe deﬁnition of b, we see that bν ∈ id(b), contradicting the fact that ν ∈ xa (A − id(x)). Since α is antitone, we obtain that bα ≤ aα, completing the proof of (16).

19.5.

Congruences

In this section we compute many properties of the congruences of A[B], to lay the groundwork for the proof of the isomorphism (4). 19.5.1

Congruence spreading

We show that congruence-perspectivities and congruence-projectivities in A naturally translate into a family of congruence-perspectivities and congruenceprojectivities in A[B]. Lemma 19.13. Let a, b, c, d ∈ A with a < b and c < d. Let p, q ∈ B with p ≤ q. If [a, b] is congruence-perspective onto [c, d] in A, then [ pab , qab ] is congruence-perspective onto [ pcd , qcd ] in A[B]. Proof. For an down congruence-perspectivity of [a, b] onto [c, d], we only have to compute four inequalities and a componentwise meet; these are trivial. For the up congruence-perspectivity, we have to verify that pab ∨ qcd = qab ; we perform the componentwise join and then apply Lemma 19.9. Since congruence-projectivity is the transitive extension of congruenceperspectivity, we obtain: Corollary 19.14. If the interval [a, b] is congruence-projective onto the interval [c, d] in A, then the interval [ pab , qab ] is congruence-projective onto the interval [ pcd , qcd ] in A[B]. If a, b ∈ A with a < b and Φ is a congruence of A[B], we deﬁne Φab as the congruence of B that corresponds to the restriction of Φ to Bab (deﬁned in Section 19.4.1) under the isomorphism ϕab : p → pab of B and Bab . The following statement easily follows from Lemma 19.13: Lemma 19.15. Let a, b, c ∈ A with a ≤ b ≤ c. Let Φ be a congruence of A[B]. Then Φac = Φab ∧ Φbc in B.

19.5. Congruences

255

Lemma 19.16. Let a, b, c, d ∈ A with a ≤ b and c ≤ d. If con(c, d) ≤ con(a, b), then Φab ≤ Φcd in B. Proof. First, let us assume that c ≺ d. Then it follows from con(c, d) ≤ con(a, b) that [a, b] is congruence-projective onto [c, d]. Let p ≤ q in B and pab , qab ] is congruencep ≡ q (Φab ). Then pab ≡ qab (Φ). By Corollary 19.14, [ projective onto [ pcd , qcd ], so pcd ≡ qcd (Φ), that is, p ≡ q (Φcd ), verifying that Φab ≤ Φcd . Second, in the general case, let c = c0 < c1 < · · · < cn = d be a maximal chain in the interval [c, d] in B. Then con(cj−1 , cj ) ≤ con(c, d) ≤ con(a, b), for every 0 < j ≤ n. Hence by the previous paragraph, Φab ≤ Φcj−1 cj . Applying Lemma 19.15 (n − 1)-times, we obtain that Φab ≤ 0<j≤n Φcj−1 cj = Φcd . Lemma 19.17. Let A be a ﬁnite simple lattice and let z be an atom of A. Let p, q ∈ B with p ≤ q and let Φ be a congruence of A[B]. Then p0A z ≡ q0A z (Φ) implies that κp ≡ κq (Φ). Proof. Since A is simple, con(0A , z) = con(0A , 1A ), so Φ0A 1A = Φ0A z , by Lemma 19.16. Therefore, if p0A z ≡ q0A z (Φ), then p ≡ q (Φ0A z ), and so p ≡ q (Φ0A 1A ), that is, p0A 1A ≡ q0A 1A (Φ), which completes the proof since κp = p0A 1A and κq = q0A 1A . For a congruence Φ of A[B], we deﬁne Φ to be the congruence of B induced by the restriction of Φ to the diagonal of A[B]. Note that Φ = Φ0A 1A . If Θ is a congruence of B, we deﬁne a binary relation ΔΘ on A[B] by α ≡ β (ΔΘ) iff aα ≡ aβ (Θ), for all a ∈ A. It is easy to verify that ΔΘ is a congruence. Lemma 19.18. For any congruence Θ of B, ΔΘ is a congruence of A[B]. Proof. It is obvious that ΔΘ is an equivalence relation. The substitution property for meets is obvious; for joins it follows from our formulas for joins in A[B]: for α, β ∈ A[B], the join is the closure of the pointwise join α ∨B A β, which is the same as p(α ∨B A β) (see Theorem 19.3). Since pa is a polynomial, for every a ∈ A, it preserves the congruence ΘA in B A . The substitution property for joins follows for ΔΘ. Lemma 19.19. ΔΦ ⊆ Φ, for any congruence Φ of A[B]. Proof. For α, β ∈ A[B], let α ≡ β (ΔΦ). By deﬁnition, aα ≡ aβ ( Φ), for any a ∈ A. Deﬁne γ = j(α) and δ = j(β). By (5), aα = (A − fil(a))γ and aβ = (A − fil(a))δ, for any a ∈ A. Also, since aα ≡ aβ (Φ), for any a ∈ A, we obtain that bγ ≡ bδ (Φ), for any b ∈ A. For b ∈ A, let b = χb ∨ κγ(b) . Then bγ, if a ∈ / id(b); b (a) = 1B , otherwise.

256

19. Tensor Extensions

Similarly, let σb = χb ∨ κδ(b) , that is, δ(b), if a ∈ / id(b); σb (a) = 1B , otherwise. Since bγ ≡ bδ (Φ) means that κγ(b) ≡ κδ(b) (Φ), for any b ∈ A, it follows that b ≡ σb (Φ), for any b. We claim that α = b∈A b . Indeed, for all a ∈ A, ab = ab b∈A

ba

(since the other components equal 1B ) = b (a) = γ(b) = ((A − fil(a))γ = aα.

b∈fil(a) /

b∈fil(a) /

Similarly, β = b∈A σb . Since b ≡ σb (Φ), for any b ∈ A, it follows that α ≡ β (Φ), completing the proof. Lemma 19.20. Let A be a ﬁnite simple lattice. Then ΔΦ = Φ, for any congruence Φ of A[B]. Proof. Let α ≡ β (Φ) with α ≤ β. If d ∈ A is join-irreducible, let c = d∗ , the unique c ∈ A covered by by d. Then, id(d) − id(c) = {d} and −−→ (α ∧ χd ) ∨ χc = α(d)cd , −−→ (β ∧ χd ) ∨ χc = β(d)cd . −−→ −−→ Therefore, α(d)bd ≡ β(d)cd (Φ), that is, dα ≡ dβ (Φcd ). By Lemma 19.16, −−→ −−→ since A is simple, dα ≡ dβ (Φ0A 1A ), that is, α(d)0A 1A = β(d)0A 1A (Φ), which is the same as κα(d) ≡ κβ(d) (Φ). By the deﬁnition of Φ, we obtain that dα ≡ dβ (Φ). If d ∈ A is join-reducible, then d = d1 ∨ · · · ∨ dn , where each di is joinirreducible. Then, by Lemma 19.1, dα = d1 α ∧ · · · ∧ dn α ≡ d1 β ∧ · · · ∧ dn β = dβ

(Φ).

So dα ≡ dβ (Φ), for any d ∈ A. Therefore, α ≡ β (ΔΦ), implying that Φ ⊆ ΔΦ. The reverse containment holds by Lemma 19.19. We have thus established the main result of this section: Theorem 19.21. Let A be a nontrivial ﬁnite simple lattice. Then A[B] is a congruence preserving extension of B. Proof. Indeed, p → κp is a congruence preserving embedding of B into A[B].

19.5. Congruences

19.5.2

257

Some structural observations

In this section we prove some easy structural results on tensor extension that will be needed in the subsequent sections. Lemma 19.22. Let B be a nontrivial bounded lattice. Then A[B] is generated, as a meet-semilattice, by the set { pa1 | a ∈ A, p ∈ B }. Proof. Let α ∈ A[B]. Let β ∈ B A satisfy α = m(β), and set −−→ (βb )b1 ∈ A[B], γ= b∈A−

where βb = bβ. Obviously, γ is meet-generated by the set { pa1 | a ∈ A− , p ∈ B }, so it suﬃces to prove that γ = α. Notice that, for a ∈ A, −−→ 1, if b ≥ a; a(βb )b1 = bβ, otherwise. Therefore, aγ =

βb = am(β) = aα,

ba

that is, γ = α. Lemma 19.23. Let us assume that 1, the unit element of A, is join-reducible. Then we can represent any α ∈ A[B] in the form α= (α ∧ χa | a ∈ D), A[B]

where D is the set of dual atoms of A. Proof. Let β = A[B] (α ∧ χa | a ∈ D). Since α ∧ χa ≤ α, for a ∈ D, the inequality β ≤ α obviously holds. Let c ∈ A − {0, 1}. Then for any dual atom a containing c, cβ ≥ cα ∧ cχa = cα holds, since c ≤ a implies that χa (c) = 1. Therefore, cβ ≥ cα, for c = 1. Since α, β ∈ hom{∨,0} (A, B d ) and 1 is join-reducible, 1β ≥ 1α also holds. Therefore, β = α, as claimed. Lemma 19.24. Let us assume that A has a unique dual atom u. Let α ≤ β in A[B]. Then (χu ∨ α) ∧ β = (χu ∧ β) ∨ α.

258

19. Tensor Extensions

See Figure 19.4. Proof. Let p = 1α. Then χu ∨ α = pu1 . Therefore, xβ, if x = 1; x((χu ∨ α) ∧ β) = p, if x = 1, since p = 1α ≤ β(1). Now

xβ, if x = 1; x(χu ∧ β) = 0B , if x = 1.

Therefore, it follows from α ≤ β that x((χu ∧ β) ∨B A

xβ, if x = 1; α) = p, if x = 1.

We conclude that (χu ∧ β) ∨B A α = (χu ∨ α) ∧ β ∈ A[B]. Therefore, (χu ∧ β) ∨B A α is closed, and (χu ∧ β) ∨ α = (χu ∧ β) ∨B A α = (χu ∨ α) ∧ β, which was to be proved.

β ∨ χu β α ∨ χu

χu

(χu ∨ α) ∧ β = (χu ∧ β) ∨ α

β ∧ χu α α ∧ χu Figure 19.4: Illustrating Lemma 19.24.

19.5. Congruences

19.5.3

259

Lifting congruences

Let Θ ∈ Con A and let α : A → B be an antitone map. Deﬁne αΘ : A/Θ → B by (17) (a/Θ)αΘ = (a/Θ)α = ( a/Θ)α. This deﬁnition makes sense since A is ﬁnite and α is antitone. It is easy to see that (α ∧ β)Θ = αΘ ∧ βΘ and (α ∨B A β)Θ = αΘ ∨B A/Θ βΘ , for antitone maps α and β. We now show that the operations of reducing maps modulo Θ and taking closures (see Section 19.3) commute. First, a simple observation. Since x ≤ x implies that xj(α) ≥ x j(α), for any α : A → B, it follows that (18) ap(α) = ( yα | x is maximal in A − fil(a) ). yx

Lemma 19.25. Let Θ ∈ Con A and let α : A → B be an antitone map. Then p(αΘ ) = p(α)Θ . Proof. By (17), it suﬃces to show that if a is minimal in a/Θ, then (a/Θ)p(αΘ ) = ap(α). If a is minimal in a/Θ, then xΘ ≥ a/Θ iff x ≥ a. Therefore, (a/Θ)p(αΘ ) = (y/Θ)αΘ xa y/Θx/Θ

=

(

(y/Θ)αΘ | x is maximal in A − fil(a) ),

y/Θx/Θ

by a variant of (18). Since a is minimal in a/Θ, both fil(a) and A − fil(a) are unions of congruence classes. Therefore, if x is maximal in A − fil(a), then x is certainly maximal in x/Θ. But if x is maximal in x/Θ, then y/Θ ≤ x/Θ iff y ≤ x. Thus (y/Θ)αΘ | x is maximal in A − fil(a) ) (a/Θ)p(αΘ ) = ( yx

(y/Θ)α | x is maximal in A − fil(a) ) = ( yx

yα | x is maximal in A − fil(a) ) = ( yx

= ap(α).

260

19. Tensor Extensions

Corollary 19.26. Let Θ ∈ Con A. Then there exists a unique homomorphism ϕ : A[B] → (A/Θ)[B] such that ϕ : pab → pa/Θ b/Θ . Proof. Let ϕ : α → αΘ . This deﬁnes a lattice homomorphism by Lemma 19.25. It is easy to check that ϕ : pab → pa/Θ b/Θ . The uniqueness follows from the fact that A[B] is generated, as a lattice, by the set { pab | a < b ∈ A }. For Θ ∈ Con A, let Θ[B] ∈ Con A[B] be the kernel of the homomorphism described in Corollary 19.26. For a < b in A, then it is apparent that ιB , if a ≡ b (Θ); (19) Θ[Bab ] = ωB , otherwise. (Bab was deﬁned in Section 19.4.1.) It is much easier to lift congruences of B to A[B] than it is to lift congruences of A. Let Φ ∈ Con B and let α : A → B. Deﬁne αΦ : A → B/Φ by composing α with the canonical projection from B to B/Φ. The following statement is trivial. Lemma 19.27. For Φ ∈ Con B, there exists a unique homomorphism λ : A[B] → A[B/Φ] −→

such that λ : pab → (p/Φ)ab . Let A[Φ] ∈ Con A[B] be the kernel of the homomorphism described in Lemma 19.27. Then (20)

A[Φ]ab = Φ,

for a < b in A. For Θ ∈ Con A and Φ ∈ Con B, Corollary 19.26 and Lemma 19.27 combine in the following commutative diagram: A[B] ⏐ ⏐

−−−−→

(A/Θ)[B] ⏐ ⏐

A[B/Φ] −−−−→ (A/Θ)[B/Φ] Let Θ[Φ] ∈ Con A[B] be the kernel of either composite. Then by (19) and (20), ιB , if a ≡ b (Θ); (21) Θ[Φ]ab = Φ, otherwise, for a < b in A.

19.5. Congruences

19.5.4

261

The main lemma

Let I be an ideal of A, and ϕ : I[B] → A[B] be the embedding of Lemma 19.11. For Φ ∈ Con A[B], let ΦI be the restriction of Φ to I[B], which is viewed as a sublattice of A[B]. Since ( pab )ϕ = pab , for a < b ∈ I, it follows that (ΦI )ab = Φab . We are now ready to prove the main lemma. Lemma 19.28. Let A be a ﬁnite lattice and let B be a bounded lattice. Let Φ, Ψ ∈ Con A[B] satisfy Φab ≤ Ψab , for all a < b ∈ A. Then Φ ≤ Ψ. Proof. We prove this statement by induction on length A. The lemma is trivial for lattices satisfying length A ≤ 2. Now let length A = n, let α ≤ β in A[B], and let α ≡ β (Φ). First, we prove Claim. Let u be a dual atom of A. Then α ∧ χu ≡ β ∧ χu

(Ψ).

Proof. Let I = id(u). Then length I < length A. For all a < b in I, by the induction hypothesis, (ΦI )ab = Φab ≤ Ψab = (ΨI )ab , therefore, ΦI ≤ ΨI . Note that α ∧ χu , β ∧ χu ∈ I[B] since, in the terminology of Lemma 19.11, α ∧ χu and β ∧ χu are “zero padded.” The congruence α ≡ β (Φ) implies that α ∧ χu ≡ β ∧ χu (Φ), so that α ∧ χu ≡ β ∧ χu (ΦI ). Since ΦI ≤ ΨI , we obtain that α ∧ χu ≡ β ∧ χu (ΨI ), that is, α ∧ χu ≡ β ∧ χu (Ψ). To prove the lemma, we consider two cases. Case 1: 1A is join-reducible. By Lemma 19.23, α = ( α ∧ χa | a ∈ D ) and β = ( β ∧ χa | a ∈ D ), where D is the set of dual atoms of A. By the Claim, α ∧ χa ≡ β ∧ χa (Ψ), for a ∈ D. Therefore, α ≡ β (Ψ). Case 2: 1A is join-irreducible. A has a unique dual atom u. Let I = id(u) and deﬁne α1 = α ∨ χu , α2 = α ∧ χu ,

β1 = β ∨ χu β2 = β ∧ χu .

Let p = 1α and q = 1β. Noting that α1 = pu1 and β1 = qu1 , it follows from Φu1 ≤ Ψu1 that α1 ≡ β1 (Ψ). By the Claim, α2 ≡ β2 (Ψ). By Lemma 19.24, α1 ∧ β = β2 ∨ α. Thus α = α ∨ α2 ≡ α ∨ β2 = α1 ∧ β ≡ β1 ∧ β = β This completes the induction.

(Ψ).

262

19. Tensor Extensions

19.6.

The congruence isomorphism

In this section we prove Theorem 19.4 in a form that describes the isomorphism. For Γ ∈ Con A[B], let F(Γ) ∈ (Con B)Con A be deﬁned by (22) F(Γ) : ΘA (ai , bi ) → Γai bi . i≤n

i≤n

It follows from Lemma 19.16 that F is a well deﬁned. By deﬁnition, F(Γ) is a {∨, 0}-homomorphism from Con A into (Con B)d . Since Con B is distributive, by Lemma 19.12, F(Γ) ∈ (Con A)[Con B]. Therefore, F : Con A[B] → (Con A)[Con B]. We show that F is an isomorphism. It is clear that F is an isotone map. It is a consequence of Lemma 19.28 that F is injective with an isotone inverse. To complete the proof, we must show that F is onto. Let η ∈ (Con A)[Con B]. Then there is ξ : Con A → Con B such that η = m(ξ). In particular, (23) con(a, b)η = con(a, b)m(ξ) = Θξ. ΘΘA (a,b)

Deﬁne Γ ∈ Con A[B] by

Γ=

con(Θ)ξ).

Θ∈Con A

Let a < b in A. Compute: con(a, b)F(Γ) = Γab =

Θ[Θξ] ab

Θ∈Con A

=

(Θ[Θξ])ab

Θ∈Con A

=

Θξ

(by (21))

Θcon(a,b)

= con(a, b)η. Since the con(a, b) {∨, 0}-generate Con A, for a < b ∈ A, and F is a {∨, 0}homomorphism, it follows that F(Γ) = η, so F is surjective. We have proved the following theorem: Theorem 19.29. Let A and B be nontrivial ﬁnite lattices. Then the map F : Con A[B] → (Con A)[Con B] is an isomorphism.

19.7. Discussion

19.7.

263

Discussion

This topic started with the 1968 paper Anderson and Kimura [1] and the 1974 paper Schmidt [106] (see also Schmidt [108] and Gr¨ atzer and Schmidt [61]), already discussed in Section 5.3. Anderson and Kimura [1] introduced tensor products (see also Fraser [14] and Shmuley [114]) as in Section 19.1. In general, for the nontrivial lattices A and B with zero, A ⊗ B is not a lattice. (For instance, M3 ⊗ F(3) is not a lattice by Gr¨ atzer and Wehrung [83].) Obviously it is if A and B are ﬁnite. For A and B ﬁnite, the lattice A[B] has a natural meet-embedding into A ⊗ B. This embedding is an isomorphism if either A or B is distributive. In general, however, |A[B]| < |A ⊗ B|. Quackenbush computed some small examples: |M3 [M3 ]| = 44 and |M3 ⊗ M3 | = 50; |N5 [N5 ]| = 42 and |N5 ⊗ N5 | = 43; |N5 ⊗ M3 | = |N5 [M3 ]| = 41. Problem 19.1. When is |A[B]| = |A ⊗ B|? When is |A[B]| = |A ⊗ B| − 1? When is |A[B]| = |A ⊗ B| − n, where n is small compared to |A| and |B|? Problem 19.2. What does the defect |A ⊗ B| − |A[B]| signify? Given |A| and |B|, what is the maximum value of the defect? Since A ⊗ B and A[B] are not the same, in general, the main result of this chapter sheds no light on the old result: (24)

Con A ⊗ Con B ∼ = Con(A ⊗ B),

of Gr¨ atzer, Lakser, and Quackenbush [43]. The isomorphisms obtained in Theorem 19.29 and in equation (24) are special cases of the main result of Gr¨atzer and Wehrung [84], using results from Gr¨ atzer and Wehrung [82] and Gr¨ atzer and Greenberg [30]. Problem 19.3. Can we utilize Theorem 19.29 to obtain a proof of the isomorphism (3)? Gr¨ atzer and Wehrung [82] and [84] present tensor product-like constructions for general lattices that guarantee that the resulting structure is a lattice: lattice tensor products in [82] and capped tensor products in [84]. For both constructions the analogues of (3) are proved. The reader is referred to Gr¨ atzer and Wehrung [86] for a survey. It is proved in Gr¨ atzer and Greenberg [29] that for nontrivial ﬁnite lattices A and B, the tensor extension A[B] is isomorphic to the lattice tensor product of A and B. (In particular, this implies the isomorphism A[B] ∼ = B[A].) So Theorem 19.29 should be viewed as an elementary proof for the ﬁnite case of the general result in Gr¨ atzer and Wehrung [84]. Problem 19.4. Find a direct proof of the isomorphism A[B] ∼ = B[A], for nontrivial ﬁnite lattices A and B?

264

19. Tensor Extensions

In the preface, we described the topic of this chapter as follows: Let be a construction for ﬁnite distributive lattices (that is, if D and E are ﬁnite distributive lattices, then D E is a ﬁnite distributive lattice). Find a construction of ﬁnite lattices (that is, if K and L are ﬁnite lattices, then K L is a ﬁnite lattice) satisfying Con(K L) ∼ = Con K Con L. There are three relevant results, for A × B, A ⊗ B, and for A[B], and in all three cases we get = . Problem 19.5. Are there lattice constructions AB for ﬁnite lattices A and B satisfying Con A Con B ∼ = Con(A B), other than A × B, A ⊗ B, and A[B]? Problem 19.6. Are there constructions A B for ﬁnite distributive lattices A and B and constructions A B for ﬁnite lattices A and B satisfying Con A Con B ∼ = Con(A B), other than those listed in Problem 19.5?

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Index

0-separating homomorphism, 17 1/3-boolean triple, 210 2-distributive lattice, 141, 226, 244, 270 2/3-boolean triple, 198 {a, b}-fork, 169 Absorption identity, 9 Adaricheva, K., xxv Algebra, 9 partial, 47 Algebraic functions, 23 Anderson, J., xxv, 263 Antichain, 4 Antisymmetric, 3 Antitone map, 4 Arber, A., xv Arguesian identity, 126 Arguesian lattice, 126 Associative identity, 9 Atom, 19 dual, 19 Atom Lemma, 53, 54, 150 Atomistic lattice, 19 Automorphism, 12 Balanced triple, 58 Baranski˘ı, V. A., xvii, 125, 177, 244, 265 Baranski˘ı-Urquhart Theorem, xvii, 244

Base congruence, 22 Berman, J., 102, 265 Bijection, 12 Binary operation, 9 Binary partial operation, 47 Binary polynomial, 23 Binary relation, 3 product, 15 Birkhoﬀ’s Subdirect Representation Theorem, 22 Birkhoﬀ, G., xviii, 22, 100, 177, 187, 265 Block, 6 trivial, 6 Bogart, K. P., 265 Bol’bot, A. D., 267 Boolean lattice, 24 Boolean triple, xvi, xx, 57, 58–70, 191, 248 1/3, 210 2/3, 198 Bounded order, 4 Breadth, 224 Capped tensor product, 263 Chain, 4 in an order, 4 Chopped lattice, xiv, xvi, 47, 48– 50, 52–55, 72, 79–83, 86, 131, 136, 137, 145, 147,

276

Index

148, 150–152, 181, 185, 186, 221, 228, 230, 231, 237 congruence, 48 ideal, 48 sectionally complemented, 53, 54, 86 Closure system, 10 CLP, xiii, 222 Colored lattice, 39, 94, 219 Coloring, 39, 94, 219 multi, 219 Commutative identity, 9 Comparable elements, 3 Compatible congruence vector, 50 Compatible vector, 49 Complement, 19 relative, 20 sectional, 20 Complemented lattice, 19 relatively, 20, 88, 140, 223, 266 sectionally, xiv, xx, 20, 47, 79, 83–88, 129, 141, 145–153, 173, 178, 180, 188, 210, 216, 227, 228, 235, 238– 241 Complete lattice, 10 Concrete category, 230 Congruence, 14, 48 base, 22 chopped lattice, 48 discrete, 15 ϕ, 190 ϕ+ , 190 forcing, 35 permutable, 15, 173, 174, 272 principal, 15 reﬂection, 39 regular, 127, 129, 173 restriction, 39 separates 0, 17 separates 1, 17 uniform, 129 Congruence class, 14 Congruence lattice, 15

distributivity of, 37 Congruence Lattice Problem, xiii, 210 Congruence permutable, 15 Congruence relation, see Congruence Congruence restriction, 14 Congruence vector, 50 compatible, 50 Congruence-determining sublattice, 43 Congruence-ﬁnite, 223 Congruence-forcing, 35 Congruence-perspective, 35 down, 35 up, 35 Congruence-preserving extension, 42 ϕ, 190 ϕ+ , 190 Congruence-preserving sublattice, 42 Congruence-projective, 35 Congruence-reﬂecting extension, 41 Congruence-reﬂecting sublattice, 41 Contravariant functor, 230 Convex embedding, 13 Convex extension, 13 Convex sublattice, 13 Convex subset, 13 Covering, 5 CPE-class, 209, 210, 216, 223 Crawley, P., xvi, 79, 150, 153, 158, 159, 265 Cubic extension, xvi, xvii, xix, 72, 74, 75, 145–148, 151–155, 158, 159, 161, 162, 178 Cycle, 8 Davey, B., xxv Davey, B. A., xv, 266, 267 Diagram, 5 Digraph, 182 labeled, 182 Dilworth Theorem, xiii, xiv, xvi, 79, 80, 85, 87, 93, 216,

Index

245 Full, 85 Dilworth, R. P., xiii, xvi, 36, 79, 100, 150, 153, 158, 159, 265, 266 Direct power, 21 Direct product, 5, 20 Discrete congruence, 15 Discrete-transitive lattice, 139 Distributive lattice, 24 Dobbertin, H., 266 Doubly 2-distributive lattice, 226 Down congruence-perspective, 35 Down-set, 4 Dual, 4 Dual atom, 19 Duality Principle for Lattices, 9 Duality Principle for Orders, 4 Edge, 178 Embedding, 12 convex, 13 ideal, 14 Endomorphism, 12 Equation, 24 Equivalence relation, 7 Erd˝ os, P., xv Extension, 13 congruence-preserving, 42 congruence-reﬂecting, 41 convex, 13 ϕ-congruence, 190 ϕ-congruence-preserving extension, 190 ϕ+ -congruence, 190 ϕ+ -congruence-preserving extension, 190 Filter, 14 principal, 14 Fraser, G. A., 263, 266 Free lattice, 26 Freese, R., 100, 102, 126, 127, 265– 267

277

Fried, E., v Frucht, R., 177–180, 186–188, 266 Fuchs, L., v Full Dilworth Theorem, 85 Funayama, N., xiii, 37, 79, 266 Functor, 230 contravariant, 230 Ganter, B., 267 Generating set (of a sublattice), 13 Glued sum, 6 Gluing, 27 repeated, 30 Gorbunov, V. A., 267 Graph, 178 rigid, 179 Greatest lower bound, 4 Greenberg, M., xvii, 190, 198, 246, 263, 267 Greferath, M., 267 Haviar, M., xxv Herrmann, C., 125, 271 Homomorphic image, 16 Homomorphism, 12 0-separating, 17 kernel, 16 Homomorphism Theorem, 16 Huhn, A. P., xiv, xvii, 215, 221, 222, 224, 226, 244, 271, 272 Ideal, 13, 48 chopped lattice, 48 principal, 14 proper, 14 Ideal embedding, 14 Ideal lattice, 14 Idempotent identity, 9 Identity, 24 absorption, 9 arguesian, 126 associative, 9 commutative, 9

278

Index

idempotent, 9 Incomparable elements, 3 Independence Theorem, 177 Strong, 177 Independence Theorem for Modular Lattices, 125 Inequality, 24 Injective, 13 Interval, 13 perspective, 32 prime, 13 projective, 33 trivial, 13 Isoform lattice, 139 Isomorphism, 4, 12 Isomorphism Theorem for Modular Lattices, 33 Isotone function, 24 Isotone map, 4 Janowitz, M. F., 88, 272 Jeˇzek, J., 102, 266 Jipsen, P., 267 Join, 9 Join-independent, 99 Join-irreducible element, 19 Join-reducible element, 19 Join-semilattice, 10 Kaarli, K., 174, 272 Katriˇ na´k, T., xxv Kelly, David, v, xxv, 103, 267 Kernel, 16 Kung, J. P. S., 265 Labeled digraph, 182 Lakser, H., v, xiv, xvi, xvii, 47, 52, 54, 79, 86, 93, 100, 102, 106, 183, 188, 216, 217, 219, 220, 222–224, 227, 228, 242, 246, 263, 267–269 Landau, E., xviii Lattice, 8, 9 2-distributive, 141, 226, 244, 270

arguesian, 126 as algebra, 9 as order, 8 atomistic, 19 boolean, 24 chopped, xiv, xvi, 47, 48–50, 52–55, 72, 79–83, 86, 131, 136, 137, 145, 147, 148, 150–152, 181, 185, 186, 221, 228, 230, 231, 237 colored, 39, 94, 219 complemented, 19 complete, 10 congruence, 15 distributivity of, 37 congruence permutable, 15 discrete-transitive, 139 distributive, 24 doubly 2-distributive, 226 Duality Principle, 9 embedding, 12 endomorphism, 12 homomorphism, 12 ideal, 14 isoform, 139 isomorphism, 12 lower semimodular, 33 modular, xvi, xix, xx, xxii, 25, 26, 28, 31–35, 39, 62, 115– 127, 141, 153, 155–157, 189, 244, 266, 270, 273 Independence Theorem, 125 n-distributive, 127, 226, 271 non-separating, 17 nontrivial, 8 planar, xvii, 10, 60, 93, 97, 106, 114, 122, 160, 224, 225, 228, 242, 243, 268 quotient, 16 regular, 127, 129, 173 relatively complemented, 20, 88, 140, 223, 266 rigid, 12 sectionally complemented, xiv

Index

sectionally complemented, xx, 20, 47, 79, 83–88, 129, 141, 145–153, 173, 178, 180, 188, 210, 216, 227, 228, 235, 238–241 semimodular, xvi, xx, 33, 57, 85, 87, 88, 105–114, 141, 153–160, 178, 188, 209, 210, 216, 244, 268, 270 separable, 163 simple, 15 subdirectly irreducible, 21 trivial, 8 uniform, xvi, xx, 127, 129, 130, 132, 136, 137, 140, 141, 161, 174, 271–273 weakly atomic, 19, 115, 116 Lattice identity, 24 Lattice inequality, 24 Lattice tensor product, 263 Least upper bound, 3 Leder, G., xxv Length, 4 Linear order, 4 Lower bound, 4 greatest, 4 Lower semimodular lattice, 33 Maeda, F., 15, 272 McKenzie, R. N., xviii, 272 McNulty, G., xviii, 272 Meet, 9 Meet-irreducible element, 19 Meet-reducible element, 19 Meet-semilattice, 10 Mendelsohn, E., 125, 272 Merging, 48 Modular lattice, xvi, xix, xx, xxii, 25, 26, 28, 31–35, 39, 62, 115–127, 141, 153, 155– 157, 189, 244, 266, 270, 273 Independence Theorem, 125 Monolith, 22

279

Morphism, 230 Multi-coloring, xvi, 219 n-distributive lattice, 127, 226, 271 Nakayama, T., xiii, 37, 79, 266 Nation, J. B., 102, 266 Natural equivalence, 231 Natural transformation, 230 Non-separating lattice, 17 Nontrivial lattice, 8 Object, 230 Operation, 9 binary, 9 partial, 47 partial binary, 47 Order, 3 associated with preorder, 8 bounded, 4 diagram, 5 dual, 4 Duality Principle, 4 length, 4 linear, 4 planar, 5 trivial, 3 Order dimension, 225 Ordering, 3 Ordinal sum, 6 Ore, O., 105, 150, 272 Osborne, Michael, xxv Partial algebra, 47 Partial operation, 47 Partition, 6, 15 lattice, 7 Permutable congruences, 15, 173, 174, 272 Perspectivity of intervals, 32 Planar lattice, xvii, 10, 60, 93, 97, 106, 114, 122, 160, 224, 225, 228, 242, 243, 268 Planar order, 5 Ploˇsˇcica, M., xxv

280

Index

Polynomial, 23 binary, 23 unary, 23 Polynomial function, 23 Power set, 4 Preorder, 7 Priestley, H. A., xv, 266, 267 Prime interval, 13 Principal congruence, 15 Principal ﬁlter, 14 Principal ideal, 14 Product direct, 20 subdirect, 21 Projection map, 20 Projectivity of intervals, 33 Proof-by-Picture, xv Proper ideal, 14 Pruning, xvi, 139 Pudl´ ak, P., xiii, 150, 158, 159, 222, 272 Pultr, A., 187, 272 Quackenbush, R. W., v, xvi, 161, 174, 246, 263, 268, 269 Quotient lattice, 16 Reﬂection, 39 Reﬂection map, 39 Reﬂexive, 3 Reﬂexive product, 30 Reﬂexive-transitive closure, 8 Regular congruence, 127, 129, 173 Regular lattice, 127, 129, 173 Relation binary, 3 reﬂexive product, 30 Relational product, 15 reﬂexive, 30 Relative complement, 20 Relatively complemented lattice, 20, 88, 140, 223, 266 Repeated gluing, 30 Restriction, 39

congruence, 14 Restriction map, 39 Reuter, K., 100, 272 Rigid, 12 Rigid graph, 179 Rival, I., xvi, 93, 225, 269 Roddy, M., 54, 86, 268, 269 Rose, H., 267 Sabidussi, G., 272 Schmidt, E. T., v, xiii, xiv, xvi, xvii, 15, 47, 53, 57, 62, 63, 65, 68, 71, 73, 79, 83, 86–90, 93, 100–103, 106, 115, 125– 127, 129, 139, 140, 145, 153, 161, 174, 177, 189, 190, 198, 207, 211, 216, 219, 220, 222–224, 226, 227, 246, 263, 266–273 Schmidt, S. E., 267 Scott, M., xxv Second Isomorphism Theorem, 16 Sectional complement, 20 Sectionally complemented chopped lattice, 53, 54, 86 Sectionally complemented lattice, xiv, xx, 20, 47, 79, 83–88, 129, 141, 145–153, 173, 178, 180, 188, 210, 216, 227, 228, 235, 238–241 Semilattice homomorphism, 12 join, 10 meet, 10 Semimodular lattice, xvi, xx, 33, 57, 85, 87, 88, 105–114, 141, 153–160, 178, 188, 209, 210, 216, 244, 268, 270 Separable lattice, 163 Separates, 17 Separator, 163, 175 Shmuley, Z., 263, 273 Simple lattice, 15 Smirnov, D. M., 267

Index

Spectrum, 89 Strong Independence Theorem, 177 Subdirect product, 21 Subdirectly irreducible lattice, 21 Sublattice, 13 congruence-determining, 43 congruence-preserving, 42 congruence-reﬂecting, 41 convex, 13 generating set, 13 Suborder, 4 Substitution Property, 14 Sum, 6 glued, 6 ordinal, 6 Surjection, 12 Surjective, 13 Symmetric, 7 Taylor, W.F., xviii, 272 Tensor extension, 248 Tensor product capped, 263 lattice, 263 Teo, S.-K., 100, 102, 273 Term, 23 Thomsen, K., xvi, 129, 161, 271 Tischendorf, M., 146, 222, 273 Transitive, 3 Triple balanced, 58 boolean, xvi, xx, 57, 58–70, 191, 248 Trivial block, 6 Trivial interval, 13 Trivial lattice, 8 Trivial order, 3

281

Trnkov´ a, V., 187, 272 T˚ uma, J., xiv, 150, 158, 159, 211, 222, 272, 273 Tumanov, V. I., 267 Unary polynomial, 23 Uniform congruence, 129 Uniform lattice, xvi, xx, 127, 129, 130, 132, 136, 137, 140, 141, 161, 174, 271–273 Unit, 4 Up congruence-perspective, 35 Upper bound, 3 least, 3 Upper semimodular lattice, 33 Urquhart, A., 125, 177, 244, 273 Valuation, 90 Variety, 22, 25 Vector compatible, 49 compatible congruence, 50 congruence, 50 Vertex, 178 Wang, D., 102, 225, 271 Weakly atomic lattice, 19, 115, 116 Wehrung, F., v, xiv, xxv, 57, 60, 102, 103, 188, 211, 217, 222, 263, 267, 269, 271, 273 Wille, R., 100, 267, 272 Wolk, B., 100, 269 Zaguia, N., xvi, 93, 225, 269 Zero, 4 Zhang, V., 102, 225, 273